# Question 1caee

Dec 11, 2017

${\text{CO}}_{2}$

#### Explanation:

The question essentially wants you to figure out the molar mass, i.e. the mass of exactly $1$ mole, of this compound.

Now, the important thing to keep in mind here is that $1$ mole of a molecular substance contains $6.022 \cdot {10}^{23}$ molecules of that substance $\to$ this is known as Avogadro's constant.

In other words, in order to have $1$ mole of a molecular substance, you need to have a sample that contains $.6022 \cdot {10}^{23}$ molecules of that substance.

This means that the molar mass of the compound will give you the mass of $6.022 \cdot {10}^{23}$ molecules.

Since you already know that $4.67 \cdot {10}^{16}$ molecules have a mass of $3.41 \cdot {10}^{- 6}$ $\text{g}$, you can use this as a conversion factor to find the mass of $6.022 \cdot {10}^{23}$ molecules.

6.022 * 10^(23) color(red)(cancel(color(black)("molecules"))) * (3.41 * 10^(-6)color(white)(.)"g")/(4.67 * 10^(16)color(red)(cancel(color(black)("molecules"))))= "44.0 g"#

You can thus say that you have $\text{44.0 g}$ for every $6.022 \cdot {10}^{23}$ molecules, i.e. $1$ mole of this compound, which means that you have

${\text{molar mass" = "44.0 g mol}}^{- 1} \to$ three sig figs

Based on the options given to you, the closest match is carbon dioxide, ${\text{CO}}_{2}$, which has a molar mass of ${\text{44.01 g mol}}^{- 1}$

Dec 11, 2017

a) $C {O}_{2}$

#### Explanation:

The mole is related to the mass by the molecular weight. We can calculate the ratio of molecules to those of a mole and compare that to the ratio of compound molecular weights to the sample mass. The one with the equal ratios is the answer.

Ratio of molecules to a Mole: $\frac{4.67 \times {10}^{16}}{6.022 \times {10}^{23}}$
$= 7.75 \times {10}^{- 8}$

$\frac{M a s s}{\text{Molecular Weight") = (3.41 xx 10^(-6))/("M.W.}}$

$C {O}_{2} = \frac{3.41 \times {10}^{- 6}}{44} = 7.75 \times {10}^{- 8}$
$C {H}_{4} = \frac{3.41 \times {10}^{- 6}}{16} = 2.13 \times {10}^{- 8}$
$N {H}_{3} = \frac{3.41 \times {10}^{- 6}}{17} = 2.01 \times {10}^{- 8}$
${H}_{2} O = \frac{3.41 \times {10}^{- 6}}{18} = 1.89 \times {10}^{- 8}$