# Chemistry

A compound containing only carbon, hydrogen and oxygen is subjected to elemental analysis. Upon complete combustion, a .1804g sample of the compound produced .3051g of CO2 and .1249g of H2O. What is the empirical formula of the compound?

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1. Someone else did a problem just like it, and here is his setup; I just changed the numbers, and you should check my numbers

mass of C in sample = mass of C in CO2
mass of C in sample = (0.3051 g CO2 / 44 g/mol)(1 mol C / 1 mol CO2)(12 g/mol) = 0.08321 g

mass of H in sample = mass of H in H2O
mass of H in sample = (0.1249 g H2O / 18 g/mol)(2 mol H / 1 mol H2O)(1 g/mol) = 0.01388 g

mass of O in sample = total mass of sample - mass of C - mass of H
mass of O in sample = 0.1804 g - 0.08321 g - 0.01388 g = 0.08691 g
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Calculate for moles of C, H & O from the calculated masses:

moles of C = 0.08321 g/ 12 g/mol = 0.006934 mol
moles of H = 0.01388 g / 1 g/mol = 0.01388 mol
moles of O = 0.08691 g / 16 g/mol = 0.005432 mol
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Divide this moles with the lowest value which is 0.006934 mol

C = 0.006934/0.006934 = 1.00
H = 0.01388/0.006934= 2.00
O = 0.005432/0.006934= 0.783
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Multiply by 4 to make the ratios a whole number:

C = 1 x 4 = 4
H = 2 x 4 = 8
O = 0.783 x 4=3.13=3

Empirical Formula is C4H8O3

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2. You are appriciated

You saved me

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3. No. That is wrong. The lowest value that you should divided is 0.005432 mol. Then you can get:

C = 0.006934/0.005432 = 1.3
H = 0.01388/0.005432 = 2.6
O = 0.005432/0.005432 = 1

C = 1.3 x 3 = 3.9 ~ 4
H = 2.6 x 3 = 7.8 ~ 8
O = 1 x 3 = 3

Only then the Empirical Formula is C4H8O3

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