# Help! Empirical/Molecular Problem!?

Help! gurus of chemistry.. question: A 64.213g sample of an unknown compound contained the elements carbon, hydrogen and nitrogen in unknown ratios. It was determined that the combustion of this compound provided 128.2319g of CO2 and 78.73713g H2O. A separate mass spectrometry experiment confirmed the molecular weight of this compound was 88.1528 amu. Use this infomation to determine the molecular formula of this compound.

I would like to know how to do it!

1) First, we know that the problem substance contianed H, C and N and that it reacted with oxygen yielding:

X + O2 ------>CO2 + H2O

We know that were produced 128.2319 g of CO2 and 78.73713g of H2O

First, how much Carbon and Hydrogen is present in products?

%C in CO2 = (12 / 44 )x100% = 27.27%
%H in H2O = (2 / 18) x100% = 11.11%

mass C in CO2 = (128.2319 g)(27.27%) = 34.968 g
mass H in H2O = (78.73731 g)(11.11%) = 8.74 g

These must be the same amounts of each element present inthe original sample of 64.213, hence, the amount of nytrogen should be:

mass of N = 64.213 - (34.968 +8.74) = 20.505 g

2) With those data, we can compute number of moles of each element in sample:

n(C) = 34.968 g / 12 g/mol = 2.91
n(H) = 8.74 g / 1 g/mol = 8.74
n(N) = 20.505 g / 14 g/mol = 1.46

We can see how is the ratio between the elements:

H / C = 8.74 / 2.91 = 3
C / N = 2.91 / 1.46 = 2
H / N = 8.74 / 1.46 = 6

Hence, the empirical formula is:

C2H6N

(molecular weight of this formula is 52

3) Knowing that molecular weight of the sample is 88.1528 uma then

88.1528 / 52 = 1.695

we can think that this amount is almost "2" in round numbers, then,

Molecular formula of compound is:

(C2H6N)2 or more likely:

C4H12N2

Hope it helps!

Good luck!
All of the carbon must be in the CO2 and all of the hydrogen must be in the water.

Using the molar ratio of carbon in CO2 (12/44), the carbon content is 34.97g

Using the molar ratio of Hydrogen in water (2/18), the hydrogen content is 8.747g

Subtracting both from the original 64.213g leaves 20.496g of nitrogen.

Now you can calculate the empirical formula in the ususal way. Divide the grams of each by the respective molecular mass and simplifying to whole numbers to get the molar ratio of C2H6N1 as the empirical formula.

Based on the molecular mass of 88g, the molecular formula is C4H12N2

Nice problem
? mol C = 128.2319 g CO2 / 44.010 = 2.9137
? g C = 2.9137 x 12.011 = 34.996

? mol H = 78.73713 g H2O / 18.02 g/mol x 2 = 8.739
? g H = 8.739 x 1.008 =8.8089 g

34.996 + 8.8089 = 43.805 g

64.213 - 43.805 = 20.408 g of N
20.408 g / 14.0067 =1.457 moles N

we have : 2.9137 moles C, 8.739 moles H and 1.457 moles N
We divide for the smallest number
2.9137/ 1.457 = 2
8.739 / 1.457 = 6
1.457 / 1.457 = 1
the empirical formula is
C2H6N ( mass = 45 g/mol)

88.1518 / 45 = 2 (about )
the molecular formula is
C4H12N2
if the combustion of 64.21g of compound gives you 128.231g of co2 then 88.1528g will give you its 176.31g according to farmula(128.23/64.21)*88.1528=... co2. now how much gm of C is in 176.31gm of co2, because this C belongs to the compound combusted. in 44g CO2, 12g C is there then in 176.31g CO2, there would be(12/44)*176.31=47.60=48g of C. and now similarly for hydrogen it would be (78.73713/64.21)*88.152=108.43... of water will be produced by 88.152g of compound on combustioning. now one mole of water contains 2g H, then in 108.43 it would be, (2/18)*108.43=12.05=12g. now 88.152g of our compound contains 12g of H + 48g of C and rest is the Nitrogen. youy can calculate the amont of nitrogen in your compound now according to farmula, 12gm H+48gm C+ xgm N= 88.152gm compound. the amount of Nitrogen is 28gm in its one mole. according to this information one molecule of this compound will have 12 atoms of H, 4 atoms of C and 2 atoms of Nitrogen. and its molecular farmula will be C4H12N2.

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