Question

i need the steps. thanks you

a-answer is ch3

muta Determine the empirical formula of a hydrocarbon that produced 15.318 H20 and 30.31g CO2 15.3la H2O

b- answer is C7H12

Answer 1

Ans :

a)

Mol C = mol CO2 = mass / molar mass

= 30.31 g / 44.01 g/mol = 0.689 mol

Mol H2O = 15.31 g / 18.01528 g/mol = 0.850 mol

Mol H = 2 x 0.850 mol = 1.70 mol

dividing the number of mol by a commom value we get the ratio as : 2 : 5

so the empirical formula of compound will be : C₂H₅

b)

Let us assume 1 mol of compound = 95 g

mass C = (87.22 x 95 g ) / 100 = 82.859 g

mol C = mass / molar mass

= 82.859 g / 12.01 g/mol = 7 mol

mass H = (12.78 x 95 g ) / 100 = 12.141 g

mol H = 12.141 / 1.008 g/mol = 12

So the molecular formula of the compound = C₇H₁₂