## test 1, question 7

Rosamari Orduna 1D
Posts: 31
Joined: Fri Apr 06, 2018 11:04 am

### test 1, question 7

Determine the empirical formula for an organic compound that, when burned in the presence of oxygen, formed 69.4g of CO2 and 17.05g of H2O. What is the mass percentage of carbon and hydrogen in the sample?

So, I was able to figure out the percentages, but how am i supposed to figure out the empirical formula when there is no overall mass given for me to subtract each element from?

Can someone guide me through this to get the answer, because I cannot figure it out...

Rebecca Chu 1C
Posts: 27
Joined: Fri Apr 06, 2018 11:02 am

### Re: test 1, question 7

An organic compound contains C and H so you know your empirical formula will be $C_{x}H_{y}$ (because we don't know what x and y are yet).
And then you are given grams of carbon dioxide which you can use to find the moles of C. You do the same thing with water (i.e. find moles of H)
Basically: 69.4 g $CO_{2}$ x (1 mol $CO_{2}$ / 44.01 g) x (1 mol C / 1 mol $CO_{2}$) = 1.58 mol C
and 17.05 g $H_{2}O$ x (1 mol $H_{2}O$ / 18.02 g) x (2 mol H / 1 mol $H_{2}O$) = 0.95 mol H
then you divide by the smallest number (which is 0.95 in this problem) since a molecule must have whole atoms
so for C, 1.58/0.95 = 1.663 and for H, 0.95/0.95 = 1.00
since C is not in whole numbers and you need whole numbers, multiply everything by 3 to get a whole number (or very very close to a whole number)
C: 1.663 x 3 = 4.989 which you can round up to 5 since it's super close and H: 1.00 x 3 = 3
So your empirical formula would be $C_{5}H_{3}$

(I had a different test version but the question asks for the same thing, just that the numbers were different so if someone can confirm that the answer is right, I'd appreciate it.)

105012653 1F
Posts: 30
Joined: Fri Apr 06, 2018 11:02 am

### Re: test 1, question 7

I had trouble with this question too... great explanation!^^