# An unknown compound is analyzed and found to be composed of 14.79% nitrogen, 50.68% oxygen, and 34.53% zinc. What is the empirical formula for the compound?

Jun 18, 2018

The empirical formula is ${\text{ZnN"_2"O}}_{6}$.

#### Explanation:

To begin, using a $\text{100-g}$ sample as a proportion to 100%, convert each percentage of elements into grams.

Your values will look like this :

• $\text{14.79 g }$ nitrogen
• $\text{50.68 g }$ oxygen
• $\text{34.53 g }$ zinc

Next, convert each value into moles by using molar mass.

(Divide your original value by the molar mass to find moles)

$\text{14.79 g N" / "14.07 g.mol" = "1.051 moles nitrogen}$

$\text{50.68 g O" / "16 g" = "3.168 moles oxygen}$

$\text{34.53 g Zn" / "65.39 g" = "0.528 moles zinc}$

Take your smallest value of moles and divide every numerical value by it. In this case, the smallest value is $0.528$. The unit of mol will be canceled out automatically by the division, only concern yourself with the numerical values.

$\frac{1.051}{0.528} = 1.99$

$\frac{3.168}{0.528} = 6$

$\frac{0.528}{0.528} = 1$

To find the smallest whole number ratio, multiply each value by $2$.

$1.99 \cdot 2 \approx 4$

$6 \cdot 2 = 12$

$1 \cdot 2 = 2$

${\text{Zn"_2"N"_4"O}}_{12}$

Remember, empirical formulas are related by a whole number ratios or the least common multiple of all numbers. $4$, $12$, and $2$ all share the factor of $2$. Divide each subscript by $2$.

The empirical formula is:

${\text{ZnN"_2"O}}_{6}$

(Remember that if the subscript of an element is $1$, then no subscript is necessary).

Jun 18, 2018

$Z n {N}_{2} {O}_{6}$

#### Explanation:

As always we ASSUME a $100 \cdot g$ mass of compound, and we assess the EMPIRICAL formula by dividing the elemental masses thru by the ATOMIC masses of each element...

$\text{Moles of zinc} \equiv \frac{34.53 \cdot g}{65.4 \cdot g \cdot m o {l}^{-} 1} = 0.528 \cdot m o l$

$\text{Moles of nitrogen} \equiv \frac{14.79 \cdot g}{14.01 \cdot g \cdot m o {l}^{-} 1} = 1.056 \cdot m o l$

$\text{Moles of oxygen} \equiv \frac{50.68 \cdot g}{15.999 \cdot g \cdot m o {l}^{-} 1} = 3.168 \cdot m o l$

And so we divide thru by the SMALLEST molar quantity to get a trial empirical formula of...

$Z {n}_{\frac{0.528 \cdot m o l}{0.528 \cdot m o l}} {N}_{\frac{1.056 \cdot m o l}{0.528 \cdot m o l}} {O}_{\frac{3.168 \cdot m o l}{0.528 \cdot m o l}} \equiv Z n {N}_{2} {O}_{6}$...(and I hope you can see those quotients....because I had to magnify them...)

And of course this is zinc nitrate...$Z n {\left(N {O}_{3}\right)}_{2}$. Note that normally you would NEVER be quoted percentage oxygen. You would be quoted percentage metal, percentage nitrogen, and then you would be expected to get percentage oxygen by difference. Oxygen is a difficult element to analyze...