Finding the mass with C1V1=C2V2?

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In summary: It becomes easy and obvious to use.In summary, In order to prepare a solution of MgO in water, you need to add 800mL of water to obtain a concentration of 1700mg/L.
  • #1
CB23ESC
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1. You need to prepare a solution of MgO in water, what mass of MgO will you need to add 800mL of water to obtain a concentration of 1700mg/L?

The equation to do this is C1V1=C2V2 (as stated by lecturer)

I am honestly so confused. Please help me!
 
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  • #2
The product of concentration and volume is mass.
1700mg/L means if you wanted to make 1 L of the solution you would need 1700mg.
If you only want to make 800mL, you add the appropriate portion.
800mL/1L = what fraction? Take the same fraction of 1700mg and you should have the same concentration.
I don't see how C1V1 = C2V2 helps in this case.
 
  • #3
I'm glad you agree that C1V1=C2V2 doesn't add up! I'm so lost... I was thinking maybe my lecturer got it wrong but I don't know. Is there any other way to find out the mass using the details in the question?
 
  • #4
Yes. proportionality.
Your target concentration is 1700mg/L, so how many mg would be in 800mL?
Concentration x Volume = mass.
 
  • #5
Okay target concentration is 1700mg/L, the mg in the 800mL would be 800g (800000mg) because it's water?
This honestly seems very odd... I wonder why he would of told me to use the other equation.

My friend thought maybe C1 = 1700, V1 = 1, C2 = 0.8 and then find V2. But the answer came up as 2125L... which also seems very out of proportion!
 
  • #6
Alright, so I've tried this using your suggestion.

C1 Is 1700
V1 is 0.8.

We're trying to achieve a concentration of 1700 with a volume of 0.8.
Multiply these and you'll get 1360, which equals C2xV2.
Since we're looking for a mass, not a concentration or volume. C2 is in mg/L and V2 is in L.
If you multiply these together the Ls cancel out and you are left with just mass. So the mass is 1360g?

Does that seem right?
 
  • #7
CB23ESC said:
Okay target concentration is 1700mg/L, the mg in the 800mL would be 800g (800000mg) because it's water?
This honestly seems very odd... I wonder why he would of told me to use the other equation.

My friend thought maybe C1 = 1700, V1 = 1, C2 = 0.8 and then find V2. But the answer came up as 2125L... which also seems very out of proportion!
I have no clue what you are doing.
Assume you have one liter (=1000mL) of solution that has 1700mg in it. If you pour out 800mL of that solution, what portion of you 1700mg would be in that 800mL?
 
  • #8
CB23ESC said:
Alright, so I've tried this using your suggestion.

C1 Is 1700
V1 is 0.8.

We're trying to achieve a concentration of 1700 with a volume of 0.8.
Multiply these and you'll get 1360, which equals C2xV2.
Since we're looking for a mass, not a concentration or volume. C2 is in mg/L and V2 is in L.
If you multiply these together the Ls cancel out and you are left with just mass. So the mass is 1360g?

Does that seem right?
Yes. That makes much more sense.
 
  • #9
Okay, Thank you so much for your help! You have no idea how much I appreciate it. Thanks again.
 
  • #10
This is a question in very starting chemistry. If this was a problem, if you needed a formula, or needed the indications of a teacher, if you were stuck by that question, and you are starting chemistry, let me suggest you do yourself a big favour:

Revise SIMPLE PROPORTIONS!

Whether it needs ten minutes and then it comes back, or whether you take a school arithmetic book and work through numbers of examples, do it till it is easy and obvious.

Because ALL the quantitative problems you will meet and be asked to solve for the first years in Chenistry will be really about nothing but proportions, all the quantitative experiments for which you will have to calculate some answers will be about proportions, the very laws of Chemistry are about proportions (and Dalton's Laws or Gay-Lussac's Law are actually called Laws of Proportions - atomic theory was based on them). Indispensible concecpts like mole and atomic and molecular masses are about proportions or ratios. And even when later they are also about something else, proportions will still be used as an essential part of almost everything, in theory, examinations, or laboratory practice.

And many students have a hard time with them. They account for about the second most frequent kind of question on this subforum. Why? Why are they stuck like you? I guess because they never learned proportions at school, but more likely they learned way back in school but have forgotten, or learned but that was called arithmetic and this is called chemistry and students don't carry stuff over between subjects. And teachers of Chemistry don't reckon to teach arithmetic all over again, they assume the students do or should know it and hence that their uses of proportions are obvious.

So if you get the idea of proportions or ratios right and familiar you should almost sail through the quantitative part of chemistry, otherwise you'll stumble and be confused or forget stuff and maybe get to hate it. So I suggest you do yourself that ^^ favor. :oldsmile:
 
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Related to Finding the mass with C1V1=C2V2?

1. What is the equation C1V1=C2V2 used for?

The equation C1V1=C2V2 is used to find the mass of a substance in a solution. It is commonly used in chemistry experiments to determine the concentration of a solute.

2. How does this equation work?

This equation is based on the principle of conservation of mass. It states that the mass of a substance in a solution remains constant, even if the volume changes. By knowing the initial and final concentrations and volumes, we can calculate the mass of the substance.

3. What do C1, V1, C2, and V2 represent in the equation?

C1 and V1 represent the initial concentration and volume of the solution, respectively. C2 and V2 represent the final concentration and volume of the solution, respectively.

4. Can this equation be used for any type of solution?

Yes, this equation can be used for any type of solution as long as the concentration and volume measurements are accurate.

5. Are there any limitations to using this equation?

One limitation of using this equation is that it assumes that the volume of the solution is the same before and after the dilution process. It also does not take into account any potential errors in measurement.

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