# A compound of hydrogen and oxygen is analyzed and a sample of the compound yields 0.590 g of hydrogen and 9.40 g of oxygen. The molecular mass of this compound is 34.00 g/mol. Find the empirical formula and the molecular formula for the compound?

Nov 22, 2017

Well, it looks likely that we got $\text{hydrogen peroxide}$, ${H}_{2} {O}_{2}$...

#### Explanation:

But we must interrogate the $\text{empirical formula}$, and then use the quoted molecular mass to calculate the $\text{molecular formula...}$

And thus for the given mass of compound we gots....

$\frac{0.590 \cdot g}{1.00794 \cdot g \cdot m o {l}^{-} 1} = 0.585 \cdot m o l$ with respect to hydrogen....

and $\frac{9.40 \cdot g}{15.00 \cdot g \cdot m o {l}^{-} 1} = 0.585 \cdot m o l$ with respect to oxygen....

And thus we get the $\text{empirical formula,}$ the simplest whole number ratio representing constituent atoms in a species of ...

${H}_{\frac{0.585 \cdot m o l}{0.585 \cdot m o l}} {O}_{\frac{0.585 \cdot m o l}{0.585 \cdot m o l}} = H O$...

But it is a fact that the $\text{molecular formula}$ is a simple whole number multiple of the $\text{empirical formula.}$.

And thus in terms of the given $\text{molecular mass....}$

$34.00 \cdot g \cdot m o {l}^{-} 1 = n \times \left\{1.00794 + 15.999\right\} \cdot g \cdot m o {l}^{-} 1$...where of course $1.00794 \cdot g \cdot m o {l}^{-} 1$ and $15.999 \cdot g \cdot m o {l}^{-} 1$ are the respective ATOMIC masses of hydrogen and oxygen....

And so $n = \frac{34.00 \cdot g \cdot m o {l}^{-} 1}{\left\{1.00794 + 15.999\right\} \cdot g \cdot m o {l}^{-} 1}$

Clealry, $n = 2$, and the $\text{molecular formula} \equiv {H}_{2} {O}_{2}$, $\text{hydrogen peroxide}$ as we anticipated.....