Ethylene glycol is used as an anti-freeze in car radiators. It contains carbon, hydrogen and oxygen. Complete oxidation of a 3.10g sample of ethylene glycol produced 4.40g of carbon dioxide and 2.70g of water.
Q. Determine the mass of oxygen in the sample.
Another way of doing this question is
1. You can find the moles of CO2, so you can find the mass of carbon in CO2. Since all the carbon in the sample is contained in the CO2 afterwards, this would be the same as the mass of Carbon in the sample.
2. You can find the moles of H2O, so you can find the mass of Hydrogen in H2O. Similar reasoning, mass is neither created or destroyed, the bits of Hydrogen in the sample isn't just going to disappear, we can assume that ALL of it will end up in the H2O. Hence this will be the same as the mass of Hydrogen in the sample.
3. The sample is made up of Carbon, Hydrogen and Oxygen. You know the total mass of the sample. You can subtract the mass of carbon and the mass of hydrogen from the total mass of the sample.
For some calculations:
n(C) = n(CO2)
n(H) = 2 * n(H2O)
n(C) in CO2 is the same as n(C) in Ethylene Glycol. This also means that m(C) is also the same (since mass = moles * molar mass).
n(H) in H2O is the same as n(H) in Ethylene Glycol. This also means that m(H) is also the same.
therefore m(O) = m(Sample) - m(C) - m(H)
Number crunching:
You can find the M.F of Ethylene glycol on the internet , or you can do it like me , by hand.
It's probably best to figure it out by hand, yeah. You're not going to have the internet to look up stuff like that in the exam, so you have to go off whatever is given in the question or given in the databook.
Q. During the production of wine by fermentation , the glucose C6H1206 in grape juice is converted to ethanol( CH3CH2OH) by the action of enzymes from yeast . Carbon dioxide is also produced.
A particular sample of wine is diluted to produce an ethanol concentration of 5.0 v/v. Given the density of ethanol is 0.784 g/mL -1 , determine the concentration of ethanol in the diluted wine sample in mol/L-.
Thanks in advance.
I dont understand how v/v and density are related.
![](https://archive.atarnotes.com/cgi-bin/mathtex.cgi?C_6H_{12}O_6 \rightarrow CH_3CH_2OH + CO_2)
![](https://archive.atarnotes.com/cgi-bin/mathtex.cgi?\rho = 0.784 \ g/mL)
Hmm, I'm guessing v/v is percent volume per volume (some type of unit of concentration). This might be an incorrect assumption.
5.0 v/v would mean 5% of the wine is ethanol. So in 100mL of wine, there would be 5mL of ethanol. In 1000 mL of wine, there would be 50mL of ethanol.
Using the density, we can figure out the mass of the ethanol in 1L of wine
We can figure out how many moles of ethanol we have here. You can figure out the molar mass of ethanol from the formula given.
![](https://archive.atarnotes.com/cgi-bin/mathtex.cgi?n = \frac{m}{M} = \frac{39.2}{46})
Since we just have 1L of solution, the concentration is going to be the same as the number of moles we have
![](https://archive.atarnotes.com/cgi-bin/mathtex.cgi?C = \frac{n}{V} = n = \frac{39.2}{46} = 0.852 \ mol/L)
Hopefully I'm not too rusty and didn't make any mistakes.