I need help with this chemistry question

[Alice de]

A compound consisting of carbon, hydrogen and oxygen has a mass of 40.85g. Analysis shows that the compound contains 10.90g of carbon and 0.90g of hydrogen. What is the percentage compositions of the compound?

[azara]

OK, they've given you the grams of carbon and hydrogen.  So, you'll need to find the grams of oxygen, which will just be 40.85-10.90-0.90 = 29.05g.

With the grams, you can work out the moles of C, H, and O using the formula n = mass/molar mass:

n(C) = 10.9/12.01 = 0.91mol

n(H) = 0.90/1.01 = 0.89mol

n(O) = 29.05/16.00 = 1.82mol

The number of moles in the whole compounds is 0.91+0.89+1.82 = 3.62mol

Now that you know the moles of each element, you can calculate the percentage composition as a whole.  % composition = moles of the element/moles of the whole substance

%C = (0.91/3.62)*100% = 25.1%

%H = (0.89/3.62)*100% = 24.6%

%O = (1.82/3.62)*100 = 50.3%

Now's the important part, which is to check your answer! This one's simple: do my three values add up to 100%? Yes, they do, so my answer is probably correct.

I hope this helped! It was good studying for my chemistry exam in two weeks

EDIT

Here are some clearer instructions for you:

1. Work out the number of moles of each element

2. Add up the number of moles to find total moles in the compound.

3. Moles of element/moles of compound, multiply by 100% to give the percentage.  That's all there is to it!

Edited November 5, 2016 by azara

[Alice de]

9 hours ago, azara said:

OK, they've given you the grams of carbon and hydrogen.  So, you'll need to find the grams of oxygen, which will just be 40.85-10.90-0.90 = 29.05g.

With the grams, you can work out the moles of C, H, and O using the formula n = mass/molar mass:

n(C) = 10.9/12.01 = 0.91mol

n(H) = 0.90/1.01 = 0.89mol

n(O) = 29.05/16.00 = 1.82mol

The number of moles in the whole compounds is 0.91+0.89+1.82 = 3.62mol

Now that you know the moles of each element, you can calculate the percentage composition as a whole.  % composition = moles of the element/moles of the whole substance

%C = (0.91/3.62)*100% = 25.1%

%H = (0.89/3.62)*100% = 24.6%

%O = (1.82/3.62)*100 = 50.3%

Now's the important part, which is to check your answer! This one's simple: do my three values add up to 100%? Yes, they do, so my answer is probably correct.

I hope this helped! It was good studying for my chemistry exam in two weeks

EDIT

Here are some clearer instructions for you:

1. Work out the number of moles of each element

2. Add up the number of moles to find total moles in the compound.

3. Moles of element/moles of compound, multiply by 100% to give the percentage.  That's all there is to it!

Thank you but I still don't understand because the formula my chemistry teacher gave me used the mass of each element and of the compound instead of the moles.

[kw0573]

17 minutes ago, Alice de said:

Thank you but I still don't understand because the formula my chemistry teacher gave me used the mass of each element and of the compound instead of the moles.

Usually they ask you to use %composition to find a molecular formula, but now you only have to find %composition. So IB will do a lot of asking for different stuff and that's fine, you just have to get used to knowing exactly what is being asked for and how to solve for it. 

So @azara found mol percentage, but I think typically IB means and implies mass percentage. In that case for example carbon is just 10.90 / 40.85 * 100% = 26.7 mass%. IB question typically will specify in the question that it is percentage composition "by mass", and rarely "by mol", so there won't be any ambiguity on the actual exam. 

[mushroom]

Basically just find the percent mass of the compound taken up by each element in the compound, so for each element in the compound use this:

(mass of element)/(total mass of compound) x 100%

So for the carbon,

10.90g / 40.85g x 100% = 26.7% of the total mass of the compound

For the hydrogen,

0.90g / 40.85g x 100% = 2.2%

 

If you wanted to find molecular formula, you would find the composition by mol. It is something you're often asked to do, so why not do it here:

 

n_C = 10.9g / 12.01g/mol = 0.91mol

n_H = 0.90g / 1.01g/mol = 0.89mol

no = 29.05g /16.00g/mol = 1.82mol

n_compound = (0.91 + 0.89 + 1.82) mol = 3.62mol

 

so you use these values to write your formula as follows:

C_0.91H_0.89O_1.82

then you divide each of them by the smallest number of moles, so the 0.89mol,

C_1.1HO_2.04

and since your numbers can be easily rounded to whole numbers, the empirical formula can be assumed to be:

CHO₂

In this case, the molecular formula would be C₂H₂O₄

 

Hope this helps!

 

[mac117]

15 minutes ago, mushroom said:

Basically just find the percent mass of the compound taken up by each element in the compound, so for each element in the compound use this:

(mass of element)/(total mass of compound) x 100%

So for the carbon,

10.90g / 40.85g x 100% = 26.7% of the total mass of the compound

For the hydrogen,

0.90g / 40.85g x 100% = 2.2%

 

If you wanted to find molecular formula, you would find the composition by mol. It is something you're often asked to do, so why not do it here:

 

n_C = 10.9g / 12.01g/mol = 0.91mol

n_H = 0.90g / 1.01g/mol = 0.89mol

no = 29.05g /16.00g/mol = 1.82mol

n_compound = (0.91 + 0.89 + 1.82) mol = 3.62mol

 

so you use these values to write your formula as follows:

C_0.91H_0.89O_1.82

then you divide each of them by the smallest number of moles, so the 0.89mol,

C_1.1HO_2.04

and since your numbers can be easily rounded to whole numbers, the empirical formula can be assumed to be:

CHO₂

In this case, the molecular formula would be C₂H₂O₄

 

Hope this helps!

 

While this is fine, you cannot really calculate the molecular formula here, since you are not told what the relative molecular formula of the compound is. Your empirical formula is as far as you can go with the information provided by the OP, since you need the ratio between the relative molecular mass and empirical formula mass to have the multiplier. 

[azara]

Sorry if I caused any confusion (thanks for the correction, @kw0573) - I assumed percent by moles, rather than percent by mass.  I believe IB usually specifies if they want "% composition by mass".  If you need to find the empirical formula, you'll need % by moles, as @mushroom showed you.

What formula did your chemistry teacher give you?