Question

An unknown compound contains only carbon, hydrogen, and oxygen (cxhyoz). combustion of 6.00 g of this compound produced 8.80 g of carbon dioxide and 3.60 g of water. part a how many moles of carbon, c, were in the original sample?

Answer 1

when we have CO2 = 8.89 & H2O = 3.6 g & the total compound = 6 g So by using the molar mass of C we can get the grams of substance:
For carbon = 8.89 g (CO2)  * 12 g/mol (C)  / 44 CO2 = 2.4 g C
and then we get the moles by using the molar mass of C :
Mol of C = 2.4 g (C) * 12 g/mol = 0.2 mol C

Answer 2

1) Answer is: 0,2 mol of carbon.
Chemical reaction: CₓHₐO₀ + O₂ → xCO₂ + a/2H₂O.
m(CO₂) = 8,80 g.
n(CO₂) = m(CO₂) ÷ M(CO₂).
n(CO₂) = 8,8 g ÷ 44 g/mol.
n(CO₂) = 0,2 mol.
From chemical reaction: n(C) : n(CO₂) = 1 : 1.
n(C) = n(CO₂).
n(C) = 0,2 mol.

2) Missing question: How many moles of hydrogen,H , were in the original sample? Express your answer numerically in moles.
Answer is: 0,4 mol of hydrogen.
Chemical reaction: CₓHₐO₀ + O₂ → xCO₂ + a/2H₂O.
m(H₂O) = 3,60 g.
n(H₂O)  = m(H₂O)  ÷ M(H₂O) .
n(CO₂) = 3,6 g ÷ 18 g/mol.
n(CO₂) = 0,2 mol.
From chemical reaction: n(H) : n(H₂O) ) = 2 : 1.
n(H) = 2n(H₂O).
n(C) = 0,4 mol.