# Question 67d87

##### 1 Answer
Jun 12, 2017

$1.2 \times {10}^{7}$ ${\text{g H}}_{2}$

#### Explanation:

We're asked to find the mass, in $\text{g}$, of a sample of hydrogen gas containing $3.5 \times {10}^{30}$ hydrogen molecules.

To convert from particles to moles, we can use Avogadro's number, $6.022 \times {10}^{23} \text{particles"/"mol}$.

After this conversion, we can use the molar mass of ${\text{H}}_{2}$, which is

overbrace(2)^("two atoms of H per molecule") xx overbrace(1.01"g"/"mol")^"molar mass of H" = color(red)(2.02"g"/"mol"

to convert from moles to grams.

(The number $1.01 \text{g"/"mol}$ is the same number as the relative atomic mass of hydrogen, $1.01$ $\text{amu}$, which on most periodic tables can be found directly beneath the element's symbol.)

Converting from molecules to moles, we have

3.5 xx 10^30cancel("molecules H"_2)((1"mol H"_2)/(6.022xx10^23cancel("molecules H"_2)))

 = color(blue)(5.8xx10^6 color(blue)("mol H"_2

Now, using the molar mass, let's convert from moles to grams:

color(red)(5.8xx10^6 cancel(color(red)("mol H"_2))((2.02"g H"_2)/(1cancel("mol H"_2)))

= color(purple)(1.2xx10^7 color(purple)("g H"_2

rounded to $2$ significant figures, the amount given in the problem.

Therefore, $3.5 \times {10}^{30}$ molecules of ${\text{H}}_{2}$ has a mass of color(purple)(1.2 xx 10^7# grams.