Thank you for taking a step back with this and showing us what the basis for the math is. To many programs just tell you the formulas and send you on your way!!
Thank you for this! Online classes can be really confusing and this really helped me understand what my teachers power point was explaining. I appreciate it!
+b belle Hi b belle. Glad I could help. Non of this material is that difficult if you just understand what you are actually doing. Let me know if you have any questions. It is easier if you email me at brad.wojcik@gmail.com
Hi Stacy, Thanks for the nice comment. When I teach this course, I tell the students they don't have to memorize any formulas, but if they do want to learn one, learn V1C1=V2C2. You can do the problems without using the formula, but it really comes in handy. -Dr. Brad
+Jennifer Fagan Hi Jennifer. Glad I could help. We have all done these types of problems in everyday life with no problem. If you have 12 lamps in your house and you need 2 bulbs for each lamp you need 24 bulbs. You go to the bulb store and look on the shelf and see that they come in boxes of 6 bulbs, so you get 4 boxes. This is the same as: "Make 12 mL of a solution of 2 mg/mL. Your stock solution is 6 mg/ mL, how many mL of stock solution do you need?"
Thanks so much, Brad. Please add some alligation problems esp. those that ask how much of drug B should be added to A given the amount of B to yield C%.
Hi Garrett, Thanks for the nice comment. Yes, once you understand that most of these problems are nothing more than multiplying your given by one or more forms of 1 to change its appearance, the course is pretty easy. Let me know if you have any questions on any specific problems. -Dr. Brad
Dr. Brad, thank you for the videos for the first time I can somewhat understand the problems. I was wondering if there is a trick to remembering the conversions from mg to g and so on, I realized I need those memorized and I'm having a hard time. Thank you!
Hi Matyuktanya, There are a lot of prefixes in the metric system, but in pharmacy the ones that we use most are milli- (one thousandth), micro-(one millionth) and kilo- (one thousand). If you learn that 1 g =1000 mg, 1 mg=1000 mcg, 1 L = 1000 mL, 1 kg = 1000 g you will be able to do 99% of all the problems. You can remember that kilo- means 1000 by thinking about a kilometer (kilo-meter). It is 1000 meters, Look at a ruler marked in millimeters and you can remember that milli- is one thousandth.
Hi Matyuktanya, Yes, you will be doing many different types of conversions. All the conversions are preformed in the same manner, you just have to have the correct conversion factor. You will have to learn that 1 tsp = 5 mL, etc. Dr. Brad
thanks you sir u make easy and work for solutions for pharmacy calculation . thanks again your hard work. u are very smart way to explain pharmacy math problems.
Dr Brad, Thank you that is very helpful, I should be able to learn these and do lots of practicing to get it down. Okay I'll see those links.. Also, a professor told me that in the pharmacy they use conversions for tbsp, oz, gallons and so on.. is that true?
Also, be sure to watch the video on unit conversions. If you learn to use conversion factors rather than moving the decimal point, you will be much less prone to error.
You can download my book, Pharmacy Calculations for Pharmacy Technicians: Solving Pharmacy Calculation Problems Without All the Silly Formulas, for free at payhip.com/b/5xVY. It goes into more detail than possible on the videos and is full of practice problems and answers.
Hi Ksenya, I am a little confused. Is it 10% w/v or 100 g in 100 mL? If it is 10% w/v that is 10%g/mL or 10 g in 100 mL of solution. It is best to go back to the basics and determine how many g of Na salicylate you have in your final solutions, then convert back to percentages. If you withdraw 10 mL of the 10 g/100 mL, you would have 1 g Na Sal in the 10 mL that you withdrew. 10 mL (10 g/100 mL) = 1 g. So you end up with 1 g in 100 mL after you add water up to 100 mL. Now convert back to %.
Hello I have also found your videos extremely helpful I am also attending pharm tech school and working my way through pharmaceutical calculations and have a question. Folic acid is commonly available as a 5mg/ml injection. You are to make a preparation that contains 50mcg/ml in a total volume of ml. How would I set this problem up using v1c1=v2c2
Hi Dr. Wojcik, I am taking a Pharmacy Tech course online. Your video's have been so helpful for me. I really appreciate how you simplify the problems by using everyday terms. Please help with this prob. The order calls for 120ml of a 10% solution. We have a 25% of the same solution. How many mls of the 25% solution will I use? Thank you, Chritine
Hi Christine. Thanks for the nice comment. There are a couple of different ways you can do this. Method 1) Lets assume it is a w/v problem. You are making 120 mL of a 10% solution. First determine how many grams you need in the 120 mL 120mL (10 g/100 mL) 12 g. Now you go shopping for your 12 g. You will get them from the 25% solution. 12 g ( 100 mL/25g) = 48 mL. If you don't know how to get 10 g/100 mL from a 10% solution, review the percent video. The other way is to use the formula V1C1=V2C2. That is volume 1 times concentration 1 = volume 2 times concentration 2. V1=120 mL C1 = 10% V2 = X C2= 25% 120 mL (10%) = X (25%) solving for X we get 48 mL. I hope this helps. Dr. Brad
hey Dr.brad i'm a pharmacy student and i always face problems in this would you please explain us how to calculate the concentration if you prepare for examplea .10gm% w/v (Na-salicylate in D. water) stock solution ( 100 gm Na salicylate and water up to 100 ml) and then withdraw 10, 20, 25, 35 ml from the stock solution transfer it to a test tube then complete the volume in each to up to 100 ml what would be these new concentrations?
Very helpful as I am currently in a pharmacy technician program, and the one thing that I seem to struggle with is when I am asked to find the active ingredient in a stock solution with a percentage or ratio strength. For example one of the questions I was having trouble with was: "Make 6 fl oz. (180mL) of a w/v stock solution so that one tablespoon (15mL) is diluted to 500mL produces a 1 : 2000 w/v solution (0.05%). How much active ingredient do you need for the stock solution? I tried many different methods and found out the answer was 3g but I don't understand.
+Emily Hill Hi Emily. Yes, 3 g is correct. Before you can start doing the calculations on a problem like this you have to visualize the problem, or draw a picture. I pictured the 500 mL with the active ingredient and thought, "Ok, all the active ingredient came from that 15 mL of stock solution. How many g were in that 15 mL? 1:2000 is 1 g in 2000 mL. 500 mL(1 g/2000 mL) = 0.25 g. Ok, so we know that the stock solution is 0.25 g/15 mL. Now, how many g are in 180 mL of the stock solution? 180 mL (0.25 g/15 mL) = 3 g. You have to work backwards on problems like this. Let me know if you have any more questions. -Brad
+Brad Wojcik Thank you for the help, I've been practicing questions like these and have found that with simpler stock solution calculations, I can easily use methods such as C1V1 = C2V2 but for ones like these finding the active ingredient or involving the aliquot I have to do two step proportions so it makes more sense. So if I understand this correctly, for a question like this, first we need to find out how much active ingredient is in the final dilution (500mL) with that final concentration of 1g/2000mL. Once we have that answer it is applied to how much is in one tablespoon (0.25g) that we add, so from there we have to find out how much of the active ingredient is in 180mL of what we need to dispense?
+Emily Hill Well, try this one, so you can be sure you understand. Make 200 mL of a w/v stock solution so that 5 ml diluted to 1000 mL produces a 0.005% solution. How much active ingredient do you need for the stock solution? No hurry, I will be gone the rest of the day. Also, if you want, you can email me at brad.wojcik@gmail.com. It is easier to respond.
Hi Dr Wojcik, can you please help me with this problem: Have 300ml of a 50% solution; 200ml is added to decrease concentration. How many active ingredient would be in 4oz of the diluted solution. Thank you so much. Christine
Let's assume it is w/v. Step 1) Calculate how many grams of active ingredient are in the 300 mL of 50%. Then you will have the same number of grams in 500 mL after you add the 200 mL. Then you will set up your problem 120 mL ( number of g in 500 mL/500 mL) = g You should get 36 g. Give it a try.
Percent strength for w/v is % g/ mL. You have g/mL now, so to add the %, just multiply by 100%. 1 g/100 mL (100%) = 1% g/mL or 1% w/v. Doing the other problems the same way you will get 2% w/v, 2.5% w/v, 3.5% w/v. Remember, start by calculating how much solute (the Na sal) you will end up with in the final solution, then change back to percent. Hope this helps and thanks for watching my videos. -Dr. Brad p.s if the stock solution was the 100 g/100 mL you calculate the same way and you would have
Hi brad watching your video I'm still confused. here is a question that I have been really stuck with and if you could explain it to me. a norepinephrine drip contains 4mg in 250 ml D5W. what is the concentration in mcg/ml?
+Arleene Perez Hi Arleen. You are starting with mg/ml and have to end up with mcg/ml, so just change the 4 mg to mcg. (4 mg/250 ml)(1000 mcg/mg) = 16 mcg/ ml. See how the units of mg cancel out?. Hope this helps.
hi dr wojcik how will i do this problem. hydro cortisone 3% ointment 80g is added to enough ointment base to make a total of 1 pound. what is the new percent strength of hydro cortisone ointment?
The easiest way to do this is W1C1=W2C2. Weight1 (concentration)1= weight2 (concentration)2 80g (3%) = 454g (x) x = 0.52% Or you can calculate the number of grams of HC in the 80%, then you will know how many grams are in the 454 g. Then change back to a %. 80g(3%)= 2.4 g Now you know you have 2.4 g in 454 g. (2.4g/454g) 100% = 0.52% Hope this helps.
for the first question you went over.... is it 4ml stock and 96ml solvent? is that correct? or reverse? i understand the workings but the answers aren't written down by the actual questions itself
+macmakeuploverrr Hi Mac. It is 4 mL of stock and 96 mL of diluent. (We are not dissolving anything, so it is a diluent, not solvent. Thanks for watching.
Hi Mr. Wojcik. I was a bit confused on what the two parts in the solution are called and where in the pharmacy would this type of question would need to come into play? I am a bit confused on the last problem using the first method (not the formula). This may seem silly but why did you put the 100 into the problem under the1 %? Thank you!
Sara Rodriguez Hi Sara. The two parts of a solution are the solute and the solvent. The solute is dissolved in the solvent. A diluent is added to make the solution weaker. I put 100% under the 1% to remove the %. There is a video on percentages. Basically, to add the % sign, multiply by 100%. 0.25 (100%)= 25%. To remove the % sign, divide by 100%. 25%/100%=0.25. You would use these calculations mostly in a compounding pharmacy.
Yes, the way I look at it is that you can either try to remember a list of twenty plus formulas and hope that your particular problem will fit one of the formulas, or you can understand what you are actually doing so you can set the problem up and solve it on your own.
Hi dr. Brad I have a problem here.. Cefepime HCl is available asa 1-g vial for reconstitution. The question is how many mL of diluent would need to be added to reconstitute to concentration of 100mg/mL?
Hi Amhie. You really need a little more information to do this calculation. It turns out that Cefepime HCl forms a solution when reconstituted and the volume of the dry drug does not factor into the calculation. So it is just 1 g, which you can change off the bat to 1000 mg (1 mL/100 mg) = 10 mL. If you look on the prescribing info you will see that if you add 10 mL to a 1 g vial the approximate concentration is 100 mg/mL, but you will actually be withdrawing 10.5 mL from the vial. I really don't think it is a good question.
Well, there are actually an infinite number of combinations of the three strengths if you wanted to use all three. I would just use 2 parts of 8% and 3 parts of 3%. One combination of all three is 1 part 8% one part 12% and 5 parts of 3%. To see how I got this, start by mixing equal parts of the 8 and 12% to make 10%.
hi mr . Brad can u help me to solve this problem; an aminophylline drip is running at 1mg/kg per hour in a 12-kg child. how much aminophylline is the child receiving per day?
+Ana Angel Hi Ana. When you have a ratio like 1mg/kg/hr, you can rewrite it as 1 mg/kg*h, which makes it easier to work with. You have to end up with mg and they are telling you that the child weighs 12 kg and the drip is running 1 day. Let's change 1 day into hours. So the givens are 24 h and 12 kg, the ratio is 1 mg/kg*h and the units of the answer are mg. (24 h)(12 kg)(1mg/kg*h)= 288 mg. h and kg both cancel out and you are left with mg.
yathatisgood You don't say how many grams of 14% sodium hypochlorite you are starting with, so lets just say you are starting with 100g. This means you would have 14g NaOCl /100 g of solution. Now calculate how many grams of solution you would have if you had 14 g NaOCl in a 2% solution. You can set it up as 14g NaOCl/x = 2% Changing the 2% to a decimal you have 14 g/NaOCl/x = 0.02. Solving for x you get 700g. You started with 100g, so you would have to add 600 g of water.
Hi Dr. Brad. I might just be over thinking this problem but I am still needing some help. Your videos have help quite a bit so I must thank you for that. So the question is "How many mL must be added to 150mL of a 25%(v/w) stock solution of sodium chloride to prepare a 0.9%(v/w) sodium chloride solution? Any help would be great
There are a couple of different ways of doing this. First way is to calculate how many grams of NaCl are in the 150 mL, then calculate how many mL of the 0.9% solution you could make from those grams. BTW, it is w/v, not v/w. 150 mL ( 25g/100mL) = 37.5 g. So you have 37.5 g of NaCl. How much 0.9% can you make with that? 37.5 g (100mL/0.9g) = 4,166.7 mL. You would have to add 4,016.7 mL to the 150 mL to get 4,166.7 mL. The other way is to use V1C1=V2C2. 150mL (25%) = x (0.9%) solving for x you get 4,166.7 mL. Then subtract the 150 mL as before. Hope this helps.
Brad Wojcik Sorry about the mix up with the w/v. Thank you for correcting me. I was over thinking it. Thanks so much for your help. Trying to figure out what the book is trying to tell you and then seeing an example of it makes a difference. The book reads a lot like stereo instructions at times. :)
Brad Wojcik Okay another one. Now tell me if I'm doing this right. The pharmacist has weighted 15g of coal tar and given it to the technician with the instructions to compound a 1% ointment. What will be the final weight of the correctly compounded prescription? Wouldn't I start with the 15g(1%g/100%g) due to the fact it's a w/w equation? It equals .15g. And from there I'm lost.
The 15 g of coal tar will be equal to 1% of the final product. You will be able to make 1500 g of oint with the 15 g of coal tar. You would mix the 15 g of coal tar with 1,485 g of oint base.
hi there, I'm studying for my ptcb exam and im stuck on this dextrose problem and I really could use some help. I have my exam date for 8/15/16... the question is.. YOU ARE PLANNING TO PREPARE 600ML OF 20% DEXTROSE SOLUTION, BY MIXING YOUR 5% AND 50% DEXTROSE SOLUTION. HOW MUCH OF EACH SOLUTION WILL BE NEEDED??
Hi Melissa. This is an alligation problem. If you watch the alligation video at th-cam.com/video/CujuVNn7ROY/w-d-xo.html it will explain how to do it. You will start the problem by drawing a box and putting 20 in the middle, 5 in the lower left corner, 50 in the upper left corner. If you want to email me at brad.wojcik@gmail.com I will show you how to do the problem. It will be pretty confusing without seeing the box and the calculations.
please help me... If a pharmacist added 12 g of azelaic acid to 50 g of an ointment containing 15% of azelaic acid, what would be the final cocentration of azelaic acid in the ointment?
Hi Yasser, First calculate how many grams of azelaic are in the 50 g of 15%. (0.15)(50g)=7.5 g. Now you are adding 12 g, so you will have a total of 19.5g of azelaic acid. The total amount of oint you have now is 62g (12g +50g). So you have 19.5g active ing/62 g oint. To change this to a %, multiply by 100%. (19.5g AI/62g oint)(100%)= 31.45%. Hope this helps.
Hi Georgegina. It doesn't look like I am going to be able to make one today, but here is an excellent slide share presentation on the subject. www.slideshare.net/layon89/chap3-calculation.
Hi Lauren, First you have to know what the units of the answer will be. This is a %w/w problem because you are weighing out the active ingredient and you are weighing the final product. For % w/w the units will be % g active ing/g final product. You are starting off with 150 mg/ act ing/3 g final product. You have to change the mg to g and add the %. To change the mg to g, you multiply by 1 g/ 1000 mg. To add the % sign, you multiply by 100%. So: 150mg AI/3 g final prod (1 g AI/1000 mg AI)(100%) = 5%. See how the mg cancel out? Hope this helps. PS. Cute baby!
+Shilpa Rivera Hi Shilpa. You are starting with 100 g (use g, not gms) of dextrose and have to end up with ml (you can write it ml or mL). The ratio is 50%, which is really 50% g/mL (watch the percentage video). 50% g/mL is 50 g/100 mL. You need mL in the answer, so you have to flip the ratio and use 100 mL/50 g. So--- 100 g (100 mL/50 g) = 200 mL. Hope this helps.
+Brad Wojcik Thank you so much!! Do you have more problems like this on different link because I have few more problems that I dont seem to figure out.
+Shilpa Rivera Here are some links to additional problems. Also, there is a link to a book I put together. www.dropbox.com/s/eiuykndf06an11m/Milliequivalent%20Exercise.docx?dl=0 www.dropbox.com/s/jg5dywv9bqi2f4e/Pharm%20120%20Handouts%20Book%20third%20ed%20Jan%202013nh.docx?dl=0 www.dropbox.com/s/oqfe4nmfuwornp1/Pharm%20120%20Week%20One%20Exercises%20combined%20with%20answers.docx?dl=0 www.dropbox.com/s/snw5jvu59e330w3/Pharm%20120%20Week%20Two%20Exercises%20combined%20with%20answersnh.docx?dl=0 www.dropbox.com/s/1bgzahdoyyafe7c/Powder%20Volume%20Exercise.docx?dl=0 www.dropbox.com/s/7hoqbtimvpo2kn8/Week%203%20Exercises%20combined%20with%20answersnh.docx?dl=0 Let me know if you have any questions. -Dr. Brad
+Brad Wojcik Thank you so much!! Any other Pharmacy tech math link, please do post them here. Dr. Brad, I really appreciate your sincerity in helping us achieve our goal. Math has been a problem for me so all these times I put pharmacy tech state board exam behind. Now that I have found you, I am so hopeful that I will pass the exam. God bless!! Shilpa
+Shilpa Rivera Hi Shilpa. Read over the material I sent. The most important things to know is that the units (mg, g, mL, gtt, etc.) are the most important things. The numbers just go along for the ride. About half of all the problems are done the same way (conversions, dosage calculations, drip rates, and others). Once you learn the basic method, everything will be easy. It is all in the book. If you have questions, just email me at brad.wojcik@gmail.com. It is easier to respond in the email. You will do great.
the batman Hi Batman. I assume you are referring to the 200 mg in the problem where we are making 100 ml of a 2 mg/ml solution. 100 ml ( 2mg/1ml)= 200 mg of active ingredient needed in the final solution, You will get those 200mg from your stock solution which has 50mg/ml. 200mg (1 ml/50mg)= 4 ml of stock solution. Hope this helps.
+ayub yare Here are the dropbox links where you can download the book and some exercises. Let me know if you have any questions. www.dropbox.com/s/eiuykndf06an11m/Milliequivalent%20Exercise.docx?dl=0 www.dropbox.com/s/jg5dywv9bqi2f4e/Pharm%20120%20Handouts%20Book%20third%20ed%20Jan%202013nh.docx?dl=0 www.dropbox.com/s/oqfe4nmfuwornp1/Pharm%20120%20Week%20One%20Exercises%20combined%20with%20answers.docx?dl=0 www.dropbox.com/s/oqfe4nmfuwornp1/Pharm%20120%20Week%20One%20Exercises%20combined%20with%20answers.docx?dl=0 www.dropbox.com/s/snw5jvu59e330w3/Pharm%20120%20Week%20Two%20Exercises%20combined%20with%20answersnh.docx?dl=0 www.dropbox.com/s/1bgzahdoyyafe7c/Powder%20Volume%20Exercise.docx?dl=0 www.dropbox.com/s/7hoqbtimvpo2kn8/Week%203%20Exercises%20combined%20with%20answersnh.docx?dl=0
Hi Sir,could you please help me with these two problems: 1) How many dm3 of 16% solution of CuSO4 (d=1.18g/cm3) we can get from 590,5 g of CuSO4 * 5H2O? 2) How many grams of H2O we need to add in 0.25dm3 of 35,2% solution of Ammonia (d=0.88g/cm3) in order to get 3% solution?
Hi Melinda. To begin with, they should tell you if the solutions are w/w or w/v. CuSO4 is usually w/v, so we will do it that way. First off you have to calculate how many grams of CuSO4 you can get from the 590.5 g of the pentahydrate form. You have to look up the atomic weights of both forms, then you can use those values to form a ratio of CuSO4/CuSO4-5H20. 590.5 g Pentahydrate (159.61 g CuSO4/249.69 g CuSO4-5H20) = 377.5 g CuSO4. Now you can calculate how many cc of 16% you can get from the 377.5 g 377.5 g ( 100 cc/16g) = 2359.4 cc, which is 2.359 dm3. Since they gave the density, they may be using 16% w/w, not w/v. You would use the ratio 100g/16g and you would end up with 2359.3 g., Now you change that to a volume. There is 1 cc/1.18 g. 2359.3 g(1cc/1.18g) = 1999.4 cc or roughly 2 dm3. Again, I believe they should have given you a little more information on this one.
2) The Ammonia solution is w/w, which they should tell you. First calculate how many grams in the 0.25dm3, 250 cc ( 0.88g/cc) = 220 g. Now calculate how many grams of Ammonia. 220g ( 35.2g/100g) = 77.44 g Ammonia. Now set up an equation 77.44 g = 0.03 (x) , where x is the total weight of the 3% solution. Solving for x you get 2580g. You have 77.44 g of Ammonia, so you have to add the difference, which is 2502.56 g of H20. Sorry that is is kind of hard to explain everything in detail in the space given.
How many milliliters of a 10% potassium solution are needed to make 180 ml of a 10 mg/ml concentration solution? The book has the answer and how to solve the problem but I don't know where the problem starts from. maybe it is how it is worded that I don't understand. 10 mg/ml =1000 mg/100 ml=1 g/100 ml= 1% ---->Where did the 1000mg/100ml come from? 1 g/100ml= x/180 ml x=(1 g X 180 ml)/100ml =1.8g needed 1.8 g/x =10 g/100 ml x=(1.8 g X 100 ml)/ 10 g = 18 ml Is this a % w/v problem or % w/w
That is the long way of doing it. First calculate how many grams of potassium you in in the 180 ml. 180 mL(10mg/mL)(1 g/1000 mL) = 1.8 g. Now calculate how many mL of the 10% solution you need to get 1.8 g. 10% is 10g/100mL (w/v) 1.8g(100mL/10g) = 18 mL. Just like you were baking cookies. You would calculate how many chocolate chips you need, then go to the store.
Hi Sir,Can I ask you again to solve these problems:1)How many cm3 of 85% H3PO4 solution(d=1,689g/cm3),needs to be taken and how many cm3 of H2O should be added in order to get 2dm3 of solution that has a concentration of 0.2 mol/dm3? 2)How many cm3 Of H20 should be added in 300cm3 of 48%H2SO4(d=1,38g/cm3) in order to get a solution of 1,61mol/dm3 and d=1,10g/cm3?
This is how I did it. First, you have to look up the molecular weight of H3PO4, which is 98g/mol. You know you need 0.2 mol/dm3 and you are making 2 dm3. This means that you will need 39.2 g of H3PO4 in the 2 dm3. 2 dm3(0.2mol/dm3)(98g/mol)=39.2 g. Now you look at where you will get your 39.2 g of H3PO4. You have 85% w/w H3PO4 with a density of 1.689g/cc. This means that there are 1.436 g/cc of H3PO4-- (0.85)(1.689 g/cm3)= 1.436 g/cm3. Now you can calculate how many cc of the 85% you will need to get 39.2 g. 39.2 g (1cc/1.436g) = 27.3 cc. The balance of the 2dm3 is H20, which is 1972.7 ml.
#2)First calculate how many g of H2SO4 are in the 300 cm3. 300cc(1.38g/cc)(0.48)= 198.72g. The molecular weight of H2SO4 is 98.08 g/mol, so calculate how many g of H2SO4 will be in 1 dm3. (98.08 g/mol)(1.61 mol/dm3)= 157.9 g/dm3. Now calculate how many cc of solution you can make with the 198.72 g that you started with in the 300cc. 198.72 g(1000cc/157.9g) = 1258.5 cc. You started with 300cc, so you have to add 958.5 cc. When I calculated the final density, I came up with 1.09 g/cc. (414.0 g from the 300cc and 958.5 g from the H20 = 1372.5g/1258.5 cc which is 1.09 g/cc)
Dear Sir, this is just great! I cannot thank you enough for helping me! Sorry for bothering you! When I look at how much you have written It really makes me uncomfortable.. So I will definitely disappear for a while!
lolo- could you please email your problems to me at brad.wojcik@gmail.com? I see you have a couple of problems, but I am having a problem trying to reply to them.
7 years later and you’re still helping me pass my classes! You’re a blessing for making these and being such a good teacher!
This was exactly what I needed. Math always makes more sense when you watch someone else do it. Thank you so much!
Hi Rebecca. Glad I could help. If you have any questions you can email me at brad.wojcik@gmail.com.
Hi Brad,
I wanted to say THANK YOU for your great videos!!!!
I'm going through these again as we hit the math portions of Pharmacy 260, thank you so much for these videos Dr. Brad!
Thank you for taking a step back with this and showing us what the basis for the math is. To many programs just tell you the formulas and send you on your way!!
Thank you soo much Brad, you made my life. I have no words to explain how thankful I am.
Glad I could help. Let me know if you have any questions.
You're a better teacher than the one I have right now. Thanks for these videos!
Thanks! Glad I could help!
Thank you for this! Online classes can be really confusing and this really helped me understand what my teachers power point was explaining. I appreciate it!
+b belle Hi b belle. Glad I could help. Non of this material is that difficult if you just understand what you are actually doing. Let me know if you have any questions. It is easier if you email me at brad.wojcik@gmail.com
Hi Stacy,
Thanks for the nice comment. When I teach this course, I tell the students they don't have to memorize any formulas, but if they do want to learn one, learn V1C1=V2C2. You can do the problems without using the formula, but it really comes in handy.
-Dr. Brad
Excellent job. Thanks for sharing valuable information.
You are a life saver! I was soooo lost and confused but V1C1=V2C2 makes things 100% easier. Thank you
Thanks Dr.Brad that was helpful!
Dr Brad, thanks again for the tips!
Thanks so much for this video! I'm trying to get a head start on a future lecture, and this is exactly what I needed!
+Jennifer Fagan Hi Jennifer. Glad I could help. We have all done these types of problems in everyday life with no problem. If you have 12 lamps in your house and you need 2 bulbs for each lamp you need 24 bulbs. You go to the bulb store and look on the shelf and see that they come in boxes of 6 bulbs, so you get 4 boxes. This is the same as: "Make 12 mL of a solution of 2 mg/mL. Your stock solution is 6 mg/ mL, how many mL of stock solution do you need?"
Thank you so much. That really helped me.
Kind regards
Steve
Hi Dent,
I am happy I could help you. Be sure to download my book from the main channel page.
Brad
Wow what a great explanation, thank you!
+Renna Sakka Hi Renna, Glad you enjoyed it. Let me know if you have any questions. You can email them to me at brad.wojcik@gmail.com.-Brad
Thanks so much, Brad.
Please add some alligation problems esp. those that ask how much of drug B should be added to A given the amount of B to yield C%.
+Cecille Boznango You are welcome. If you have a specific problem you would like help with, please email me at brad.wojcik@gmail.com.
Thanks Igan!
Hi Chris,
Glad they help you. Let me know if you need any help on any problems.
Hi Garrett,
Thanks for the nice comment. Yes, once you understand that most of these problems are nothing more than multiplying your given by one or more forms of 1 to change its appearance, the course is pretty easy. Let me know if you have any questions on any specific problems.
-Dr. Brad
Thankyou sir.... finally I understood....I am greatful to you 😊
Glad I could help!
thank you. l really appreciate that you are helping me.
+Ana Angel I am always glad to help.
Dr. Brad, thank you for the videos for the first time I can somewhat understand the problems. I was wondering if there is a trick to remembering the conversions from mg to g and so on, I realized I need those memorized and I'm having a hard time. Thank you!
Hi Matyuktanya,
There are a lot of prefixes in the metric system, but in pharmacy the ones that we use most are milli- (one thousandth), micro-(one millionth) and kilo- (one thousand). If you learn that 1 g =1000 mg, 1 mg=1000 mcg, 1 L = 1000 mL, 1 kg = 1000 g you will be able to do 99% of all the problems. You can remember that kilo- means 1000 by thinking about a kilometer (kilo-meter). It is 1000 meters, Look at a ruler marked in millimeters and you can remember that milli- is one thousandth.
Hi Matyuktanya,
Yes, you will be doing many different types of conversions. All the conversions are preformed in the same manner, you just have to have the correct conversion factor. You will have to learn that 1 tsp = 5 mL, etc.
Dr. Brad
You are awesome dr.brad;) igan;)
thanks you sir u make easy and work for solutions for pharmacy calculation . thanks again your hard work. u are very smart way to explain pharmacy math problems.
Thanks for the nice comment. Let me know if you have any questions.
Dr Brad,
Thank you that is very helpful, I should be able to learn these and do lots of practicing to get it down. Okay I'll see those links..
Also, a professor told me that in the pharmacy they use conversions for tbsp, oz, gallons and so on.. is that true?
Also, be sure to watch the video on unit conversions. If you learn to use conversion factors rather than moving the decimal point, you will be much less prone to error.
You can download my book, Pharmacy Calculations for Pharmacy Technicians: Solving Pharmacy Calculation Problems Without All the Silly Formulas, for free at payhip.com/b/5xVY. It goes into more detail than possible on the videos and is full of practice problems and answers.
Hi Ksenya, I am a little confused. Is it 10% w/v or 100 g in 100 mL? If it is 10% w/v that is 10%g/mL or 10 g in 100 mL of solution. It is best to go back to the basics and determine how many g of Na salicylate you have in your final solutions, then convert back to percentages. If you withdraw 10 mL of the 10 g/100 mL, you would have 1 g Na Sal in the 10 mL that you withdrew. 10 mL (10 g/100 mL) = 1 g. So you end up with 1 g in 100 mL after you add water up to 100 mL. Now convert back to %.
Hello I have also found your videos extremely helpful I am also attending pharm tech school and working my way through pharmaceutical calculations and have a question. Folic acid is commonly available as a 5mg/ml injection. You are to make a preparation that contains 50mcg/ml in a total volume of ml. How would I set this problem up using v1c1=v2c2
What is the total volume of the 50 mcg/mL?
Hi Dr. Wojcik, I am taking a Pharmacy Tech course online. Your video's have been so helpful for me. I really appreciate how you simplify the problems by using everyday terms. Please help with this prob.
The order calls for 120ml of a 10% solution. We have a 25% of the same solution. How many mls of the 25% solution will I use? Thank you, Chritine
Hi Christine. Thanks for the nice comment. There are a couple of different ways you can do this. Method 1) Lets assume it is a w/v problem. You are making 120 mL of a 10% solution. First determine how many grams you need in the 120 mL 120mL (10 g/100 mL) 12 g. Now you go shopping for your 12 g. You will get them from the 25% solution. 12 g ( 100 mL/25g) = 48 mL. If you don't know how to get 10 g/100 mL from a 10% solution, review the percent video. The other way is to use the formula V1C1=V2C2. That is volume 1 times concentration 1 = volume 2 times concentration 2. V1=120 mL C1 = 10% V2 = X C2= 25% 120 mL (10%) = X (25%) solving for X we get 48 mL. I hope this helps.
Dr. Brad
hey Dr.brad i'm a pharmacy student and i always face problems in this would you please explain us how to calculate the concentration if you prepare for examplea .10gm% w/v (Na-salicylate in D. water) stock solution ( 100 gm Na salicylate and water up to 100 ml) and then withdraw 10, 20, 25, 35 ml from the stock solution transfer it to a test tube then complete the volume in each to up to 100 ml what would be these new concentrations?
Very helpful as I am currently in a pharmacy technician program, and the one thing that I seem to struggle with is when I am asked to find the active ingredient in a stock solution with a percentage or ratio strength. For example one of the questions I was having trouble with was: "Make 6 fl oz. (180mL) of a w/v stock solution so that one tablespoon (15mL) is diluted to 500mL produces a 1 : 2000 w/v solution (0.05%). How much active ingredient do you need for the stock solution?
I tried many different methods and found out the answer was 3g but I don't understand.
+Emily Hill Hi Emily. Yes, 3 g is correct. Before you can start doing the calculations on a problem like this you have to visualize the problem, or draw a picture. I pictured the 500 mL with the active ingredient and thought, "Ok, all the active ingredient came from that 15 mL of stock solution. How many g were in that 15 mL? 1:2000 is 1 g in 2000 mL. 500 mL(1 g/2000 mL) = 0.25 g. Ok, so we know that the stock solution is 0.25 g/15 mL. Now, how many g are in 180 mL of the stock solution? 180 mL (0.25 g/15 mL) = 3 g. You have to work backwards on problems like this. Let me know if you have any more questions.
-Brad
+Brad Wojcik Thank you for the help, I've been practicing questions like these and have found that with simpler stock solution calculations, I can easily use methods such as C1V1 = C2V2 but for ones like these finding the active ingredient or involving the aliquot I have to do two step proportions so it makes more sense.
So if I understand this correctly, for a question like this, first we need to find out how much active ingredient is in the final dilution (500mL) with that final concentration of 1g/2000mL. Once we have that answer it is applied to how much is in one tablespoon (0.25g) that we add, so from there we have to find out how much of the active ingredient is in 180mL of what we need to dispense?
Yes, that is correct.
Brad Wojcik Thank you, I think I understand it much better now.
+Emily Hill Well, try this one, so you can be sure you understand. Make 200 mL of a w/v stock solution so that 5 ml diluted to 1000 mL produces a 0.005% solution. How much active ingredient do you need for the stock solution? No hurry, I will be gone the rest of the day. Also, if you want, you can email me at brad.wojcik@gmail.com. It is easier to respond.
Hi Dr Wojcik, can you please help me with this problem: Have 300ml of a 50% solution; 200ml is added to decrease concentration. How many active ingredient would be in 4oz of the diluted solution. Thank you so much. Christine
Let's assume it is w/v. Step 1) Calculate how many grams of active ingredient are in the 300 mL of 50%. Then you will have the same number of grams in 500 mL after you add the 200 mL. Then you will set up your problem 120 mL ( number of g in 500 mL/500 mL) = g
You should get 36 g. Give it a try.
Percent strength for w/v is % g/ mL. You have g/mL now, so to add the %, just multiply by 100%. 1 g/100 mL (100%) = 1% g/mL or 1% w/v. Doing the other problems the same way you will get 2% w/v, 2.5% w/v, 3.5% w/v. Remember, start by calculating how much solute (the Na sal) you will end up with in the final solution, then change back to percent.
Hope this helps and thanks for watching my videos.
-Dr. Brad
p.s if the stock solution was the 100 g/100 mL you calculate the same way and you would have
Brad Wojcik Thank you very much from KSA
Hi brad watching your video I'm still confused. here is a question that I have been really stuck with and if you could explain it to me.
a norepinephrine drip contains 4mg in 250 ml D5W. what is the concentration in mcg/ml?
+Arleene Perez Hi Arleen. You are starting with mg/ml and have to end up with mcg/ml, so just change the 4 mg to mcg. (4 mg/250 ml)(1000 mcg/mg) = 16 mcg/ ml. See how the units of mg cancel out?. Hope this helps.
hi dr wojcik how will i do this problem.
hydro cortisone 3% ointment 80g is added to enough ointment base to make a total of 1 pound. what is the new percent strength of hydro cortisone ointment?
The easiest way to do this is W1C1=W2C2. Weight1 (concentration)1= weight2 (concentration)2
80g (3%) = 454g (x) x = 0.52% Or you can calculate the number of grams of HC in the 80%, then you will know how many grams are in the 454 g. Then change back to a %. 80g(3%)= 2.4 g Now you know you have 2.4 g in 454 g. (2.4g/454g) 100% = 0.52% Hope this helps.
yes it do thanks
I have posted links to some written material under channel posts. You should download the handout book.
for the first question you went over.... is it 4ml stock and 96ml solvent? is that correct? or reverse? i understand the workings but the answers aren't written down by the actual questions itself
+macmakeuploverrr Hi Mac. It is 4 mL of stock and 96 mL of diluent. (We are not dissolving anything, so it is a diluent, not solvent. Thanks for watching.
Set it up like this: 120 mL (10 mg/mL) = V2 (10 g/100 mL) then solve for V2. You should get 12 mL
Hi Mr. Wojcik. I was a bit confused on what the two parts in the solution are called and where in the pharmacy would this type of question would need to come into play? I am a bit confused on the last problem using the first method (not the formula). This may seem silly but why did you put the 100 into the problem under the1 %?
Thank you!
Sara Rodriguez Hi Sara. The two parts of a solution are the solute and the solvent. The solute is dissolved in the solvent. A diluent is added to make the solution weaker. I put 100% under the 1% to remove the %. There is a video on percentages. Basically, to add the % sign, multiply by 100%. 0.25 (100%)= 25%. To remove the % sign, divide by 100%. 25%/100%=0.25. You would use these calculations mostly in a compounding pharmacy.
Brad Wojcik Thank you so much!
How many milliliters of a 1:200 stock solution should be used to make
60 ml of a 0.025% solution?
Can you solve a question?
V1 (1 g/200 mL) = 60 mL (0.025 g/100 mL). Solving for V1 you get 3 mL.
So remembering rules instead of formulas works better in real pharmacy situations. Okay! I will have to watch this one again.
Yes, the way I look at it is that you can either try to remember a list of twenty plus formulas and hope that your particular problem will fit one of the formulas, or you can understand what you are actually doing so you can set the problem up and solve it on your own.
I am still stuck on tar and ointment calculation. how did you come up with the answer.
What is the question?
Hi dr. Brad I have a problem here.. Cefepime HCl is available asa 1-g vial for reconstitution. The question is how many mL of diluent would need to be added to reconstitute to concentration of 100mg/mL?
Hi Amhie. You really need a little more information to do this calculation. It turns out that Cefepime HCl forms a solution when reconstituted and the volume of the dry drug does not factor into the calculation. So it is just 1 g, which you can change off the bat to 1000 mg (1 mL/100 mg) = 10 mL. If you look on the prescribing info you will see that if you add 10 mL to a 1 g vial the approximate concentration is 100 mg/mL, but you will actually be withdrawing 10.5 mL from the vial. I really don't think it is a good question.
Brad Wojcik I got my question from my workbook, it’s kind a tricky for me to do these calculations..
If you have more questions you can email me at brad.wojcik@gmail.com
in what ratio 12%, 8% and 3% are mixed to get 5% alcohol........Sir give answer
Well, there are actually an infinite number of combinations of the three strengths if you wanted to use all three. I would just use 2 parts of 8% and 3 parts of 3%. One combination of all three is 1 part 8% one part 12% and 5 parts of 3%. To see how I got this, start by mixing equal parts of the 8 and 12% to make 10%.
hi mr . Brad can u help me to solve this problem; an aminophylline drip is running at 1mg/kg per hour in a 12-kg child. how much aminophylline is the child receiving per day?
+Ana Angel Hi Ana. When you have a ratio like 1mg/kg/hr, you can rewrite it as 1 mg/kg*h, which makes it easier to work with. You have to end up with mg and they are telling you that the child weighs 12 kg and the drip is running 1 day. Let's change 1 day into hours. So the givens are 24 h and 12 kg, the ratio is 1 mg/kg*h and the units of the answer are mg. (24 h)(12 kg)(1mg/kg*h)= 288 mg. h and kg both cancel out and you are left with mg.
If I have 14% sodium hypochlorite m/m
How can I dilute it to make 2% solution?
yathatisgood You don't say how many grams of 14% sodium hypochlorite you are starting with, so lets just say you are starting with 100g. This means you would have 14g NaOCl /100 g of solution. Now calculate how many grams of solution you would have if you had 14 g NaOCl in a 2% solution. You can set it up as 14g NaOCl/x = 2% Changing the 2% to a decimal you have 14 g/NaOCl/x = 0.02. Solving for x you get 700g. You started with 100g, so you would have to add 600 g of water.
I posted some exercises and my handout book on the main page. I think it will help you if you download them and do the exercises.
Hi Dr. Brad. I might just be over thinking this problem but I am still needing some help. Your videos have help quite a bit so I must thank you for that. So the question is "How many mL must be added to 150mL of a 25%(v/w) stock solution of sodium chloride to prepare a 0.9%(v/w) sodium chloride solution?
Any help would be great
There are a couple of different ways of doing this. First way is to calculate how many grams of NaCl are in the 150 mL, then calculate how many mL of the 0.9% solution you could make from those grams. BTW, it is w/v, not v/w. 150 mL ( 25g/100mL) = 37.5 g. So you have 37.5 g of NaCl. How much 0.9% can you make with that? 37.5 g (100mL/0.9g) = 4,166.7 mL. You would have to add 4,016.7 mL to the 150 mL to get 4,166.7 mL. The other way is to use V1C1=V2C2. 150mL (25%) = x (0.9%) solving for x you get 4,166.7 mL. Then subtract the 150 mL as before. Hope this helps.
Brad Wojcik Sorry about the mix up with the w/v. Thank you for correcting me. I was over thinking it. Thanks so much for your help. Trying to figure out what the book is trying to tell you and then seeing an example of it makes a difference. The book reads a lot like stereo instructions at times. :)
You are welcome. Most books I have seen on pharmacy calculations are very confusing. Let me know if you need any more help.
Brad Wojcik Okay another one. Now tell me if I'm doing this right. The pharmacist has weighted 15g of coal tar and given it to the technician with the instructions to compound a 1% ointment. What will be the final weight of the correctly compounded prescription? Wouldn't I start with the 15g(1%g/100%g) due to the fact it's a w/w equation? It equals .15g. And from there I'm lost.
The 15 g of coal tar will be equal to 1% of the final product. You will be able to make 1500 g of oint with the 15 g of coal tar. You would mix the 15 g of coal tar with 1,485 g of oint base.
hi there, I'm studying for my ptcb exam and im stuck on this dextrose problem and I really could use some help. I have my exam date for 8/15/16... the question is..
YOU ARE PLANNING TO PREPARE 600ML OF 20% DEXTROSE SOLUTION, BY MIXING YOUR 5% AND 50% DEXTROSE SOLUTION. HOW MUCH OF EACH SOLUTION WILL BE NEEDED??
Hi Melissa. This is an alligation problem. If you watch the alligation video at th-cam.com/video/CujuVNn7ROY/w-d-xo.html it will explain how to do it. You will start the problem by drawing a box and putting 20 in the middle, 5 in the lower left corner, 50 in the upper left corner. If you want to email me at brad.wojcik@gmail.com I will show you how to do the problem. It will be pretty confusing without seeing the box and the calculations.
please help me...
If a pharmacist added 12 g of azelaic acid
to 50 g of an ointment containing 15% of azelaic acid, what would be the final cocentration of azelaic acid in the ointment?
Hi Yasser,
First calculate how many grams of azelaic are in the 50 g of 15%. (0.15)(50g)=7.5 g. Now you are adding 12 g, so you will have a total of 19.5g of azelaic acid. The total amount of oint you have now is 62g (12g +50g). So you have 19.5g active ing/62 g oint. To change this to a %, multiply by 100%. (19.5g AI/62g oint)(100%)= 31.45%. Hope this helps.
Brad Wojcik thank you .....this helped me too much
Hi Brad
Would you be able to do a demonstration video on aliquots?
I will try to make one tomorrow. If you have a problem you would like help on, just email it to me at brad.wojcik@gmail.com
Hi Georgegina. It doesn't look like I am going to be able to make one today, but here is an excellent slide share presentation on the subject. www.slideshare.net/layon89/chap3-calculation.
Thank you Brad, that helps a lot :)
Georgie
How many milliliters of a 1:50 stock solution should be used to prepare
one liter of a 1:4000 solution
???????????
1000 mL (1 g/4000 mL) = V2(1 g/50 mL). V2 = 12.5 mL
@@BradWojcikPharmD Thank y.ou
How do you solve?
3g of Zovirax ointment contains 150mg of acyclovir. What is the concentration of this product?
Hi Lauren, First you have to know what the units of the answer will be. This is a %w/w problem because you are weighing out the active ingredient and you are weighing the final product. For % w/w the units will be % g active ing/g final product. You are starting off with 150 mg/ act ing/3 g final product. You have to change the mg to g and add the %. To change the mg to g, you multiply by 1 g/ 1000 mg. To add the % sign, you multiply by 100%. So: 150mg AI/3 g final prod (1 g AI/1000 mg AI)(100%) = 5%. See how the mg cancel out? Hope this helps. PS. Cute baby!
If 100 gms of dextrose is ordered using 50% dextrose solution, how many ml are needed? Thanks a bunch Dr. Brad!
+Shilpa Rivera Hi Shilpa. You are starting with 100 g (use g, not gms) of dextrose and have to end up with ml (you can write it ml or mL). The ratio is 50%, which is really 50% g/mL (watch the percentage video). 50% g/mL is 50 g/100 mL. You need mL in the answer, so you have to flip the ratio and use 100 mL/50 g. So--- 100 g (100 mL/50 g) = 200 mL. Hope this helps.
+Brad Wojcik Thank you so much!! Do you have more problems like this on different link because I have few more problems that I dont seem to figure out.
+Shilpa Rivera Here are some links to additional problems. Also, there is a link to a book I put together. www.dropbox.com/s/eiuykndf06an11m/Milliequivalent%20Exercise.docx?dl=0
www.dropbox.com/s/jg5dywv9bqi2f4e/Pharm%20120%20Handouts%20Book%20third%20ed%20Jan%202013nh.docx?dl=0
www.dropbox.com/s/oqfe4nmfuwornp1/Pharm%20120%20Week%20One%20Exercises%20combined%20with%20answers.docx?dl=0
www.dropbox.com/s/snw5jvu59e330w3/Pharm%20120%20Week%20Two%20Exercises%20combined%20with%20answersnh.docx?dl=0
www.dropbox.com/s/1bgzahdoyyafe7c/Powder%20Volume%20Exercise.docx?dl=0
www.dropbox.com/s/7hoqbtimvpo2kn8/Week%203%20Exercises%20combined%20with%20answersnh.docx?dl=0
Let me know if you have any questions.
-Dr. Brad
+Brad Wojcik Thank you so much!! Any other Pharmacy tech math link, please do post them here. Dr. Brad, I really appreciate your sincerity in helping us achieve our goal. Math has been a problem for me so all these times I put pharmacy tech state board exam behind. Now that I have found you, I am so hopeful that I will pass the exam. God bless!! Shilpa
+Shilpa Rivera Hi Shilpa. Read over the material I sent. The most important things to know is that the units (mg, g, mL, gtt, etc.) are the most important things. The numbers just go along for the ride. About half of all the problems are done the same way (conversions, dosage calculations, drip rates, and others). Once you learn the basic method, everything will be easy. It is all in the book. If you have questions, just email me at brad.wojcik@gmail.com. It is easier to respond in the email. You will do great.
Where did we get the 200?
the batman Hi Batman. I assume you are referring to the 200 mg in the problem where we are making 100 ml of a 2 mg/ml solution. 100 ml ( 2mg/1ml)= 200 mg of active ingredient needed in the final solution, You will get those 200mg from your stock solution which has 50mg/ml. 200mg (1 ml/50mg)= 4 ml of stock solution. Hope this helps.
hi doctor
what is your webside.are the books free.
+ayub yare Here are the dropbox links where you can download the book and some exercises. Let me know if you have any questions.
www.dropbox.com/s/eiuykndf06an11m/Milliequivalent%20Exercise.docx?dl=0
www.dropbox.com/s/jg5dywv9bqi2f4e/Pharm%20120%20Handouts%20Book%20third%20ed%20Jan%202013nh.docx?dl=0
www.dropbox.com/s/oqfe4nmfuwornp1/Pharm%20120%20Week%20One%20Exercises%20combined%20with%20answers.docx?dl=0
www.dropbox.com/s/oqfe4nmfuwornp1/Pharm%20120%20Week%20One%20Exercises%20combined%20with%20answers.docx?dl=0
www.dropbox.com/s/snw5jvu59e330w3/Pharm%20120%20Week%20Two%20Exercises%20combined%20with%20answersnh.docx?dl=0
www.dropbox.com/s/1bgzahdoyyafe7c/Powder%20Volume%20Exercise.docx?dl=0
www.dropbox.com/s/7hoqbtimvpo2kn8/Week%203%20Exercises%20combined%20with%20answersnh.docx?dl=0
10% w/v solution if you withdrew 10 mL and q.s to 100 mL, 20% w/v, 25% w/v, 35% w/v.
Hi Sir,could you please help me with these two problems: 1) How many dm3 of 16% solution of CuSO4 (d=1.18g/cm3) we can get from 590,5 g of CuSO4 * 5H2O?
2) How many grams of H2O we need to add in 0.25dm3 of 35,2% solution of Ammonia (d=0.88g/cm3) in order to get 3% solution?
Hi Melinda. To begin with, they should tell you if the solutions are w/w or w/v. CuSO4 is usually w/v, so we will do it that way. First off you have to calculate how many grams of CuSO4 you can get from the 590.5 g of the pentahydrate form. You have to look up the atomic weights of both forms, then you can use those values to form a ratio of CuSO4/CuSO4-5H20. 590.5 g Pentahydrate (159.61 g CuSO4/249.69 g CuSO4-5H20) = 377.5 g CuSO4. Now you can calculate how many cc of 16% you can get from the 377.5 g 377.5 g ( 100 cc/16g) = 2359.4 cc, which is 2.359 dm3. Since they gave the density, they may be using 16% w/w, not w/v. You would use the ratio 100g/16g and you would end up with 2359.3 g., Now you change that to a volume. There is 1 cc/1.18 g. 2359.3 g(1cc/1.18g) = 1999.4 cc or roughly 2 dm3. Again, I believe they should have given you a little more information on this one.
2) The Ammonia solution is w/w, which they should tell you. First calculate how many grams in the 0.25dm3, 250 cc ( 0.88g/cc) = 220 g. Now calculate how many grams of Ammonia. 220g ( 35.2g/100g) = 77.44 g Ammonia. Now set up an equation 77.44 g = 0.03 (x) , where x is the total weight of the 3% solution. Solving for x you get 2580g. You have 77.44 g of Ammonia, so you have to add the difference, which is 2502.56 g of H20. Sorry that is is kind of hard to explain everything in detail in the space given.
Thank you so much! You are amazing! It really means a lot to me!
Glad I could help.
C1V1 = C2V2. boom.
Not sure what your comment means. That was covered in the video.
How many milliliters of a 10% potassium solution are needed to make 180 ml of a 10 mg/ml concentration solution?
The book has the answer and how to solve the problem but I don't know where the problem starts from. maybe it is how it is worded that I don't understand.
10 mg/ml =1000 mg/100 ml=1 g/100 ml= 1%
---->Where did the 1000mg/100ml come from?
1 g/100ml= x/180 ml
x=(1 g X 180 ml)/100ml =1.8g needed
1.8 g/x =10 g/100 ml
x=(1.8 g X 100 ml)/ 10 g = 18 ml
Is this a % w/v problem or % w/w
That is the long way of doing it. First calculate how many grams of potassium you in in the 180 ml. 180 mL(10mg/mL)(1 g/1000 mL) = 1.8 g. Now calculate how many mL of the 10% solution you need to get 1.8 g. 10% is 10g/100mL (w/v) 1.8g(100mL/10g) = 18 mL. Just like you were baking cookies. You would calculate how many chocolate chips you need, then go to the store.
Hi Sir,Can I ask you again to solve these problems:1)How many cm3 of 85% H3PO4 solution(d=1,689g/cm3),needs to be taken and how many cm3 of H2O should be added in order to get 2dm3 of solution that has a concentration of 0.2 mol/dm3?
2)How many cm3 Of H20 should be added in 300cm3 of 48%H2SO4(d=1,38g/cm3) in order to get a solution of 1,61mol/dm3 and d=1,10g/cm3?
This is how I did it. First, you have to look up the molecular weight of H3PO4, which is 98g/mol. You know you need 0.2 mol/dm3 and you are making 2 dm3. This means that you will need 39.2 g of H3PO4 in the 2 dm3. 2 dm3(0.2mol/dm3)(98g/mol)=39.2 g. Now you look at where you will get your 39.2 g of H3PO4. You have 85% w/w H3PO4 with a density of 1.689g/cc. This means that there are 1.436 g/cc of H3PO4-- (0.85)(1.689 g/cm3)= 1.436 g/cm3. Now you can calculate how many cc of the 85% you will need to get 39.2 g. 39.2 g (1cc/1.436g) = 27.3 cc. The balance of the 2dm3 is H20, which is 1972.7 ml.
#2)First calculate how many g of H2SO4 are in the 300 cm3. 300cc(1.38g/cc)(0.48)= 198.72g. The molecular weight of H2SO4 is 98.08 g/mol, so calculate how many g of H2SO4 will be in 1 dm3. (98.08 g/mol)(1.61 mol/dm3)= 157.9 g/dm3. Now calculate how many cc of solution you can make with the 198.72 g that you started with in the 300cc. 198.72 g(1000cc/157.9g) = 1258.5 cc. You started with 300cc, so you have to add 958.5 cc. When I calculated the final density, I came up with 1.09 g/cc. (414.0 g from the 300cc and 958.5 g from the H20 = 1372.5g/1258.5 cc which is 1.09 g/cc)
Dear Sir, this is just great!
I cannot thank you enough for helping me!
Sorry for bothering you!
When I look at how much you have written It really makes me uncomfortable..
So I will definitely disappear for a while!
Hi melinda, I am happy to help.
I’m Brad Wojcik!! Not you!
lolo- could you please email your problems to me at brad.wojcik@gmail.com? I see you have a couple of problems, but I am having a problem trying to reply to them.