Question

1. 0.0958 g of a compound (molar mass = 312 g/mol) are burned in oxygen gas to give 0.263 g co₂ and 0.110 g h₂o. what is its empirical formula?

step 1: calculate mass of c from mass of co₂.
? g c = g co₂ x 1 mol co₂ / molar mass co₂ x 1 mol c / 1 mol co₂ x atomic mass c
step 2: calculate mass of h from mass of h₂o. (hint: 1 mol h₂o = 2 mol h, 1 mol h₂o = 1 mol o)
? g h = g h₂o x 1 mol h₂o / molar mass h₂o x 2 mol h / 1 mol h₂o x atomic mass h
step 3: calculate the mass of o.
? g o = g methanol - g c - g h
step 4: calculate the moles of c, h, and o.
mol c = g c / atomic mass c
mol h = g h / atomic mass h
mol o = g o / atomic mass o
step 5: calculate the mole ratio to determine the subscripts. mole ratio = mole / smallest mole
answer key empirical formula: c₅₀ h₉₅ o₅ = c₁₀ h₂₀ o
step 6: find the molecular formula of the compound. n = molar mass of the molecular formula / molar mass of the empirical formula
finding the molecular formula
the molar mass of acetic acid is 146 g/mol. what is the molecular formula of acetic acid?
146 / 73 = 2 (c₃ h₆ o₃) x 2 = c₆ h₁₂ o₆

1. multiply the empirical formula by this number to get the molecular formula.

Explanation:

Step1: Calculate mass of C from mass of CO₂

Given mass of CO₂ = 0.209 g. Molar mass of CO₂ is 44 g/mol and atomic mass of C is 12 g/mol.
$g\ C=0.209\ g\ CO_2\times\frac{1\ mol\ CO_2}{44\ g\ CO_2}\times\frac{1\ mol\ C}{1\ mol\ CO_2}\times\frac{12\ g\ C}{1\ mol\ C}\approx0.057\ g\ C$

Step2: Calculate mass of H from mass of H₂O

Given mass of H₂O = 0.110 g. Molar mass of H₂O is 18 g/mol and atomic mass of H is 1 g/mol. Since 1 mol H₂O has 2 mol H.
$g\ H = 0.110\ g\ H_2O\times\frac{1\ mol\ H_2O}{18\ g\ H_2O}\times\frac{2\ mol\ H}{1\ mol\ H_2O}\times\frac{1\ g\ H}{1\ mol\ H}\approx0.012\ g\ H$

Step3: Calculate mass of O

Given mass of compound is 0.095 g.
$g\ O=0.095\ g-(0.057\ g + 0.012\ g)=0.026\ g\ O$

Step4: Calculate moles of C, H and O

$mol\ C=\frac{0.057\ g}{12\ g/mol}\approx0.00475\ mol$
$mol\ H=\frac{0.012\ g}{1\ g/mol}=0.012\ mol$
$mol\ O=\frac{0.026\ g}{16\ g/mol}\approx0.001625\ mol$

Step5: Calculate mole - ratio to determine sub - scripts

Smallest mole is 0.001625 mol.
Mole ratio of C: $\frac{0.00475\ mol}{0.001625\ mol}\approx3$
Mole ratio of H: $\frac{0.012\ mol}{0.001625\ mol}\approx7.4$ (round to 7)
Mole ratio of O: $\frac{0.001625\ mol}{0.001625\ mol}=1$
Empirical formula is C₃H₇O

Step6: Find the molecular formula

Given molar mass of compound is 312 g/mol. Molar mass of empirical formula C₃H₇O is $3\times12 + 7\times1+16=57\ g/mol$
$n=\frac{312\ g/mol}{57\ g/mol}\approx5.47$ (round to 6)
Molecular formula is (C₃H₇O)×6 = C₁₈H₄₂O₆

Answer:

Empirical formula: C₃H₇O, Molecular formula: C₁₈H₄₂O₆