Chemistry: Mass to Mass Problem

In summary: The problem with making heavy weather of these calculations is that students can make errors in their calculations by imagining the atoms, molecules and their reactions in a very concrete way - drawing little diagrams with colors for types of atom which have different weights.
  • #1
AnomalyCoder
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0

Homework Statement



CH4(g) + 3cl2(g) ----> CHCl3(g) + 3HCL(g)
How many grams of CH4 is needed to produce 50.0 g CHCL3 ?

For this problem, I'll assume Carbon = 12 amu, Hydrogen = 1 amu, and CL = 35.5 amu.

The Attempt at a Solution


See the attachment (A PDF file), for my attempt @ the solution. :cry:
I think I may have the correct answer; however, I am not positive.
 

Attachments

  • 1-Subject Notebook 3 p. 49.pdf
    102.2 KB · Views: 750
Last edited:
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  • #2
ok first off, convert your grams of CHCl3 to moles of CHCl3... you should be able to do that. Then from your chemical formula, you are able to find the ratio needed to form a certain reactant. For example, 1 mole of CH4 will form 1 mole of CHCl3. The coefficient of each molecule will tell you how much you need to form the other molecule, if you find a certain amount in moles of CHCl3 then you will need the same amount of CH4 because of the 1:1 ratio. Think of it as the ratio between jam, cheese and bread when making a simple sandwich, you need 1 slice of cheese with 1 slice of jam to be able to make a sandwich (of course you will need 2 slices of bread), now if you have 9 sandwiches at the end, that means you used 9 slices of cheese, from the 1:1 ratio you assume that you must have used 9 slices of jam and from the 1:2 ratio between the cheese and the bread, you assume that you used 18 slices of bread to make 9 sandwiches.. so going back to the problem, once you find moles of CHCl3, you go and use the ratio from the coefficients, then from that you find moles of CH4, finally you can easily go from moles of CH4 to grams of CH4... good luck
 
  • #3
hyddro said:
ok first off, convert your grams of CHCl3 to moles of CHCl3... you should be able to do that. Then from your chemical formula, you are able to find the ratio needed to form a certain reactant. For example, 1 mole of CH4 will form 1 mole of CHCl3. The coefficient of each molecule will tell you how much you need to form the other molecule, if you find a certain amount in moles of CHCl3 then you will need the same amount of CH4 because of the 1:1 ratio. Think of it as the ratio between jam, cheese and bread when making a simple sandwich, you need 1 slice of cheese with 1 slice of jam to be able to make a sandwich (of course you will need 2 slices of bread), now if you have 9 sandwiches at the end, that means you used 9 slices of cheese, from the 1:1 ratio you assume that you must have used 9 slices of jam and from the 1:2 ratio between the cheese and the bread, you assume that you used 18 slices of bread to make 9 sandwiches.. so going back to the problem, once you find moles of CHCl3, you go and use the ratio from the coefficients, then from that you find moles of CH4, finally you can easily go from moles of CH4 to grams of CH4... good luck
So was my attempt incorrect? The PDF I attached shows my work.
 
  • #4
it seems correct to me (the procedure) i don't understand your numbers tho, but it seems that you are following what i said so that must be right..
 
  • #5
hyddro said:
it seems correct to me (the procedure) i don't understand your numbers tho, but it seems that you are following what i said so that must be right..

Which numbers are unclear?
I think I have the procedure down, I want someone to verify the solution if possible though. :shy:
 
  • #6
yeah that seems correct, i am getting the same thing (6.67 g of CH4)
 
  • #7
hyddro said:
yeah that seems correct, i am getting the same thing (6.67 g of CH4)

Excellent, thank you! :approve:
 
  • #8
no problem bro, good luck
 
  • #9
I almost agree with your result but I think you have illegitimately lopped the last significant figure off the calculation, you are given 35.5 for amu of Cl there are 3 Cl atoms so you cannot finish with a whole number denominator.

Much more important than that is not to make heavy weather as many students do of these calculations. You can do that by imagining the atoms, molecules and their reactions in a very concrete way - draw little diagrams with colours for types of atom which have different weights. Then you see (12 + 4 X 1) g of CH4 goes to make (12 + 1 + 3 X 35.5) g of of CHCl3. After that is is only a question of proportions for what you need to get 50g of the later. Admittedly simple proportions have not been grasped confidently by all students.

In this case reaction is in a sense is simple, the C atoms in reactant go only to C atoms in product, in other cases that might not be so, so you have to watch it.
 
  • #10
epenguin said:
I almost agree with your result but I think you have illegitimately lopped the last significant figure off the calculation, you are given 35.5 for amu of Cl there are 3 Cl atoms so you cannot finish with a whole number denominator.

Much more important than that is not to make heavy weather as many students do of these calculations. You can do that by imagining the atoms, molecules and their reactions in a very concrete way - draw little diagrams with colours for types of atom which have different weights. Then you see (12 + 4 X 1) g of CH4 goes to make (12 + 1 + 3 X 35.5) g of of CHCl3. After that is is only a question of proportions for what you need to get 50g of the later. Admittedly simple proportions have not been grasped confidently by all students.

In this case reaction is in a sense is simple, the C atoms in reactant go only to C atoms in product, in other cases that might not be so, so you have to watch it.

That's true, I was less worried about the sig. figs here, I just wanted to make sure I grasped the concept. It's actually 119.5, I just rounded to 120 to make it easier.

I never would have thought of the problem in that way, I'll be sure to try to spot such things in the future. :smile:
 

1. How do you solve mass to mass problems in chemistry?

Mass to mass problems in chemistry involve converting the given mass of one substance to the mass of another substance in a chemical reaction. To solve these problems, you need to first balance the chemical equation and then use the molar ratios to convert the given mass to moles. Finally, use the molar mass of the desired substance to convert moles back to mass.

2. What is the molar ratio in a mass to mass problem?

Molar ratio refers to the ratio of the number of moles of one substance to another substance in a chemical equation. This ratio is used to convert between moles of different substances in a chemical reaction and is essential in solving mass to mass problems in chemistry.

3. How do you calculate the molar mass of a substance?

Molar mass is the mass of one mole of a substance and is expressed in grams per mole. To calculate the molar mass of a substance, you need to add the atomic masses of all the elements present in the molecule. This can be found on the periodic table.

4. What is the difference between mass and molar mass?

Mass refers to the amount of matter contained in an object, while molar mass refers to the mass of one mole of a substance. Molar mass is expressed in grams per mole and is a conversion factor used in many chemical calculations.

5. Can you convert between mass and moles without using the molar mass?

No, the molar mass is necessary to convert between mass and moles in chemistry. It is a conversion factor that relates the mass of a substance to its number of moles and is essential in solving mass to mass problems and other chemical calculations.

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