Combustion analysis of naphthelene, a hydrocarbon used in mothballs, produces 13.20g co2 and 1.80g h2o. calculate the empirical formula of nafthelene.

C₃H₂

Explanation:

1) Since you are told naphthelene is a hydrocarbon, you know that its chemical formula only has atoms of C and H.

2) Since combustion adds O but not C nor H, you know that all the C and H present in the products come from the naphthelene sample.

3) Calculate the amount of C in 13.20 grams of CO₂:

Use atomic masses:

C: 12.01 g/mol
O: 16.00 g/mol

⇒ molar mass of CO₂ = 12.01 g/mol + 2×16.01 g/mol = 44.01 g/mol

Proportion of C in 13.20 g of CO₂

12.01 g C / 44.01 g CO₂ = x / 13.20 g CO₂ ⇒ 

x = 13.20 g CO₂ × 12.01 g C / 44.01 g CO₂ = 3.602 g C

3) Calculate the mass of H in 1.80 g of H₂O

Atomic masses:

H: 1.01 g/mol
O: 16.0 g/mol

Molar mass H₂O = 18.02 g/mol

Proportion of H in 1.80 g H₂O

2.02 g H / 18.02 g H₂O = x / 1.80 g H₂O

x = 1.80 gH₂O × 2.02 g H / 18.02 gH₂O = 0.202 g H.

4) Calculate the number of moles of C and H in the masses calculated above

Number of moles of C = 3.602 g C / 12.01 g/mol = 0.3 mol

Number of moles of H = 0.202 g  / 1.01 g/mol = 0.2 mol

5) Divide both amounts by the smallest one to calculate the proportion in moles:

C: 0.3 / 0.2 = 1.5 mol

H = 0.2 / 0.2 = 1 mol

Multiply by 2 to obtain integers:

C: 3
H: 2

6)  Then the proportion leads to the empirical formula C₃H₂

Where are the questions

Nonmetals usually react by gaining electrons