Test #1 Question 7 mass percentage

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Nabeeha Khan 1D
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Joined: Fri Apr 06, 2018 11:03 am

Test #1 Question 7 mass percentage

Postby Nabeeha Khan 1D » Sun May 06, 2018 4:19 pm

Question 7: Determine the empirical formula for an organic compound that when burned in the presence of oxygen formed 69.4g CO2 and 17.05g of H2O. What is the mass percentage of carbon and hydrogen in the sample?

Does anyone have the correct answers for this problem?

Sara Veerman-1H
Posts: 31
Joined: Fri Apr 06, 2018 11:05 am

Re: Test #1 Question 7 mass percentage

Postby Sara Veerman-1H » Mon May 07, 2018 4:28 pm

Yeah, this is a tricky question because the amount of O2 that reacted was not given. First I divided both amounts of CO2 and H2O formed by their respective molar masses and found that they were equal to each other (both created 0.178 moles). Using this knowledge and the fact that O2 was present I created the formula x + O2 --> H2O + CO2. Balancing the equation gave me CH2O + O2 --> H2O + CO2. So the empirical formula for the sample is CH2O. In order to find the mass percentage I assumed 1.00 moles of the sample was used, so there is 1 mol C, 2 mol H, and 1 mol O. Dividing those numbers by their molar mass will give you: 12.01 g C, 16.00 g O, and 2.00 g H. Then I just divided these masses by the total mass of the sample (the total mass is the moles of the sample divided by its molar mass) and I was able to get the percentages. I'm not sure if I did this right. I remember my TA saying something along the lines that not enough information was provided and free points were given and its probably only because of that I got full points on this particular problem... But I hope this helps

Madeleine Farrington 1B
Posts: 32
Joined: Fri Apr 06, 2018 11:02 am

Re: Test #1 Question 7 mass percentage

Postby Madeleine Farrington 1B » Wed May 09, 2018 10:24 am

I had a slightly different test I think, and the values given for mine were 8.45g CO2 and 1.73g H2O. When I divided each by their respective molar masses, I got 0.192 mol CO2 and 0.096 mol H2O. Since these values are not equal but .192/.096 = 2, does that mean that in the molecular formula for my problem you would have 2CO2 + H2O for the products side?

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