# Question 06412

May 19, 2017

${C}_{3} {H}_{8}$

#### Explanation:

First, we need to find the number of moles of each substance in the hydrocarbon.

Using $n = \frac{m}{M}$:

$n \left(C\right) = \left(\frac{72}{12}\right)$ $\text{mol}$

$n \left(C\right) = 6$ $\text{mol}$

$\mathmr{and}$

$n \left(H\right) = \left(\frac{16}{1}\right)$ $\text{mol}$

$n \left(H\right) = 16$ $\text{mol}$

Then, we need to determine the simplest whole number ratio of moles of the elements present.

This can be done by dividing the mole quantities by the smaller amount:

$R i g h t a r r o w \frac{H}{C} = \frac{16}{6}$

$R i g h t a r r o w \frac{H}{C} = \frac{\frac{16}{6}}{\frac{6}{6}}$

$R i g h t a r r o w \frac{H}{C} = \frac{2.67}{1}$

After division, the values should be close enough to whole numbers to approximate.

In this case, $2.67$ is not close enough to $3.00$ to round it off.

So we must multiply both values by some number to reach a reasonable level of accuracy:

$R i g h t a r r o w \frac{H}{C} = \frac{2.67}{1} \times \frac{3}{3}$

$R i g h t a r r o w \frac{H}{C} = \frac{8.01}{3}$

We have now reached an accurate enough value to round to a whole number:

$R i g h t a r r o w \frac{H}{C} \approx \frac{8}{3}$

$\therefore \text{Empirical formula} = {C}_{3} {H}_{8}$

Therefore, the simplest formula for this hydrocarbon is ${C}_{3} {H}_{8}$.

May 19, 2017

The empirical formula for this hydrocarbon is $\text{C"_3"H"_8}$.

#### Explanation:

The simplest formula of a compound is its empirical formula, which represents the lowest whole number ratio of the elements in the compound.

You have been given the masses of carbon and hydrogen in a sample of a hydrocarbon. There are several steps to follow. Convert the masses to moles. Divide the moles by the lowest number of moles to get the subscripts. If you don't get whole numbers, then you will need to multiply the moles by a whole number that results in whole numbers.

color(blue)("Determining Moles"
Multiply the given mass of each element by the inverse of its molar mass, which is its atomic weight on the periodic table in g/mol.

$\text{C} :$$72 \text{g C"xx(1"mol C")/(12.011"g C")="5.99 mol}$$\approx \text{6 mol C}$

$\text{H} :$$16 \text{g H"xx(1"mol H")/(1.008"g H")="15.9 mol}$$\approx \text{16 mol H}$

color(blue)("Determining Subscripts for Empirical Formula"
Divide each number of moles by the lowest number of moles. This will give you the subscripts for each element in the formula.

$\text{C} :$(6color(red)cancel(color(black)("mol")))/(6color(red)cancel(color(black)("mol")))="1

$\text{H} :$(16color(red)cancel(color(black)("mol")))/(6color(red)cancel(color(black)("mol")))="2.67"#

$2.67$ is not close enough to $3$ to round to $3$. You have to multiply both numbers by a number that will make both elements have a whole number subscript. When the decimal portion of a subscript is .67, multiply both subscripts by $3$.

$\text{C} :$$1 \times 3 = 3$

$\text{H} :$$2.67 \times 3 = 8.01 \approx 8$

The empirical formula for this hydrocarbon is $\text{C"_3"H"_8}$.