# How would you determine the empirical and molecular formula for hydrogen peroxide, which is 5.94% H and 94.06% O and has a molar mass of 34.01 g/mol?

##### 2 Answers
Nov 21, 2015

The molecular formula is $\text{H"_2"O"_2}$.

#### Explanation:

Since the percentages add up to 100%, we can assume that we have a 100-g sample, which will allow us to convert the percentages into grams.

$\text{H} :$5.94%=>"5.94 g"
$\text{O} :$94.06%=>"94.06 g"

Determine Moles of Each Element

First we need to determine moles of H and O by dividing their given masses by their molar masses (atomic weight on the periodic table) in g/mol.

$\text{H} :$$5.94 \cancel{\text{g H"xx(1"mol H")/(1.00794cancel"g H")="5.89 mol H}}$

$\text{O} :$$94.06 \text{g O"xx(1"mol O")/(15.999"g O")="5.88 mol O}$

Determine Mole ratios and Empirical Formula

Since the number of moles for H and O are equal, the mole ratios are 1.

The empirical formula is $\text{HO}$

Determine the molecular formula.

The empirical formula mass is (1xx1.00794"g/mol")+(1xx15.999"g/mol")="17.007 g/mol"

The molecular formula mass is $\text{34.01 g/mol}$. To get the multiplication factor, divide the molecular formula mass by the empirical formula mass.

$\text{Multiplication factor"=(34.01"g/mol")/(17.007"g/mol")="2.000}$

To get the molecular formula, multiply the subscripts of the empirical formula times 2.

The molecular formula is $\text{H"_2"O"_2}$.

Nov 21, 2015

these formulas are used in solving the problem
n(empirical formula)=molecular formula
n=molecular formula mass/empirical formula mass

#### Explanation:

Atomic mass of H=1.008
Atomic mass of O=16
amount of hydrogen present in sample=5.94/1.008=5.8
amount of oxygen present in sample=94.06/16=5.8
RATIO: H : O
5.8 : 5.8
1 : 1
so empirical formula= HO
n(empirical formula)=molecular formula
n=molecular formula mass/empirical formula mass
n=34.01/17
=2
2(HO)=H2O2
H2O2 is molecular formula.