A compound consisting of C, H and O only, has a molar mass of 331.5g/mol. Combustion of 0.1000g of this compound caused a 0.2921g increase in the mass

In combustion analysis, the amounts of carbon dioxide and waterproduced are determined by measuring the mass increase in thecarbon dioxide and water absorbers, respectively. Typicallythe absorbers are chemicals that absorb or react with carbondioxide or water. For example, a carbon dioxide absorber maybe NaOH (which reacts with carbon dioxide to form ${\displaystyle NaHC{O}_3}$). The water absorber may be ${\displaystyle M{g\left(Cl{O}_4\right)}_2}$, a dessicant (drying agent). Because there was a mass increase of the ${\displaystyle C{O}_2}$ absorber of 0.2921 g and a mass increase of 0.0951g of the water absorber, thenthe amount of ${\displaystyle C{O}_2}$ produced from the combustion is 0.2921g and the amount of ${\displaystyle H_{2}O}$ produced is 0.0951 g.
To determine the empirical formula, calculate the grams of C from ${\displaystyle C{O}_2}$, the grams of H from ${\displaystyle H_{2}O}$, and the amountof O will be determined by taking the total mass of the substancecombusted (0.1000g) and subtracting the amount of C and H. Therefore:
gram ${\displaystyle C=\left(0.2921gC{O}_2\right)\left(\frac{1mo\le }{44g}\right)\left(\frac{1mo\le C}{mo\le C{O}_2}\right)\left(\frac{12gC}{1mo\le C}\right)=0.07966gC}$
gram ${\displaystyle H=\left(0.0951g{H}_{2}O\right)\left(\frac{1mo\le H_{2}O}{18g{H}_{2}O}\right)\left(\frac{2mo\le H_{2}O}{mo\le H_{2}O}\right)\left(\frac{1.008gH}{mo\le H}\right)=0.01065gH}$
gram ${\displaystyle O=0.1000g-(0.07966g+0.01065g)=9.69\times {10}^{-3}g}$
moles ${\displaystyle O=(9.69\times {10}^{-3}g)\left(\frac{1mo\le O}{16gO}\right)=6.056\times {10}^{-4}mo\le O}$
Calculate the moles of H and C by dividing by their respective molecular weights to find:
${\displaystyle mo\le sH=0.01057mo\le H}$
${\displaystyle mo\le C=6.638\times {10}^{-3}mo\le C}$
As the number of moles of O is the smallest, divide mole C, mole Hand mole O by mole O to obtain:
mole C/mole O = 10.96
mole H/mole O = 17.45
mole O/ mole O = 1
So have ${\displaystyle C_{10.96}{H}_{17.45}O}$ must be whole numbersso multiply coefficients by 2 to obtain the empirical formula of ${\displaystyle C_{22}{H}_{35}{O}_2}$
As the empirical formula weight is about 331 g/mole and the actualmolecular weight is 331.5 g/mole, the empirical formula of thiscompound is also the molecular formula.