# An organic compound contains 45.90% C; 2.75% H; 7.65% N; 17.50% S; and the balance oxygen. What is the empirical formula of the compound?

Feb 21, 2017

The empirical formula is ${C}_{7} {H}_{5} {O}_{3} S N$.

#### Explanation:

1. First, list all of your elements. In this case they are carbon, hydrogen, oxygen, sulphur and nitrogen.

2. Assume that you have 100g of this compound. Consequently, you can assume that you have in grams the percentage you are given for each element. For example, you can assume that you have 45.9g of carbon, 2.75g of hydrogen, and so on.

3. Now, you get out your periodic table and look for the ${A}_{r}$ of each element. For carbon it's 12.01, hydrogen is 1.01 and so on.

4. Find the number of moles you have for of each element. You do this by dividing the mass in grams by the ${A}_{r}$ for each element. So, we have:
$\frac{45.9}{12.01} = 3.82$ moles of carbon,

$\frac{2.75}{1.01} = 2.72$ moles of hydrogen,

$\frac{26.2}{16} = 1.64$ moles of oxygen,

$\frac{17.5}{32.07} = 0.55$ moles of sulphur,

$\frac{7.65}{14} = 0.55$ moles of nitrogen.

1. Lastly, you just have to divide all of the values we just calculated by the smallest number of moles that there was. So in this case the smallest is 0.55. I have rounded the numbers to integers because they have to be integers in chemical compounds. For instance, you can't have 2.5 atoms.

Carbon: $\frac{3.82}{0.55} = 7$

Hydrogen: $\frac{2.72}{0.55} = 5$

Oxygen: $\frac{1.64}{0.55} = 3$

Sulphur: $\frac{0.55}{0.55} = 1$

Nitrogen: $\frac{0.55}{0.55} = 1$

So the empirical formula is ${C}_{7} {H}_{5} {O}_{3} S N$.

Feb 21, 2017

We get an empirical formula of ${C}_{7} {H}_{5} N {O}_{3} S$. And the empirical formula represents the simplest whole number ratio that relates constituent atoms in a species.

#### Explanation:

As with all these problems, we assume a mass of $100 \cdot g$, and work out the molar quantities of each element from the given percentages in this mass, and normalize the result:

$\text{Moles of carbon} = \frac{45.90 \cdot g}{12.011 \cdot g \cdot m o {l}^{-} 1} = 3.82 \cdot m o l .$

$\text{Moles of hydrogen} = \frac{2.75 \cdot g}{1.00794 \cdot g \cdot m o {l}^{-} 1} = 2.73 \cdot m o l .$

$\text{Moles of nitrogen} = \frac{7.65 \cdot g}{14.01 \cdot g \cdot m o {l}^{-} 1} = 0.546 \cdot m o l .$

$\text{Moles of sulfur} = \frac{17.50 \cdot g}{32.06 \cdot g \cdot m o {l}^{-} 1} = 0.546 \cdot m o l .$

$\text{Moles of oxygen} = \frac{26.20 \cdot g}{15.999 \cdot g \cdot m o {l}^{-} 1} = 1.64 \cdot m o l .$

And now we divide thru by the smallest molar quantity, that of nitrogen, and sulfur. We do this to normalize the result:

$\text{Carbon atoms} = \frac{3.82 \cdot m o l}{0.546 \cdot m o l} = 7.00$

$\text{Hydrogen atoms} = \frac{2.73 \cdot m o l}{0.546 \cdot m o l} = 5.00$

$\text{Nitrogen atoms} = \frac{0.546 \cdot m o l}{0.546 \cdot m o {l}^{-} 1} = 1.00$

$\text{Moles of sulfur} = \frac{0.546 \cdot m o l}{0.546 \cdot m o {l}^{-} 1} = 1.00$

$\text{Moles of oxygen} = \frac{1.64 \cdot m o l}{0.546 \cdot m o {l}^{-} 1} = 3.00$

Note that normally we would not be given a percentage oxygen content (why not? because oxygen is experimentally hard to measure); we would calculate the %O from the difference of the other percentages from 100%.

And thus we get an empirical formula of ${C}_{7} {H}_{5} N {O}_{3} S$.

Note the $\text{molecular formula}$ is a simple whole number multiple of the $\text{empirical formula}$. We need a $\text{molecular mass determination}$ before we make this assessment. For completeness, we know that the molecular weight of saccharin is $183.15 \cdot g \cdot m o {l}^{-} 1$.

The $\text{molecular formula}$ is ALWAYS a whole number multiple of the $\text{empirical formula}$.

Thus$\text{.......................................................}$

$183.15 \cdot g \cdot m o {l}^{-} 1 = n \times \left\{7 \times 12.011 + 5 \times 1.00794 + 14.01 + 32.06 + 3 \times 15.999\right\} \cdot g \cdot m o {l}^{-} 1 = 183.15 \cdot g \cdot m o {l}^{-} 1 \times n$.

Clearly $n = 1$, and the empirical formula is the same as the molecular formula, i.e. ${C}_{7} {H}_{5} N {O}_{3} S$.

Good question. I am stealing it for my A2 class.