# A sample of propane gas contains 9.36xx10^24*"hydrogen atoms". What mass with respect to carbon, and hydrogen, AND propane, does this quantity represent?

Aug 30, 2017

You gots propane, i.e. ${C}_{3} {H}_{8}$.

#### Explanation:

Now if I specify a mole of propane I specify $6.022 \times {10}^{23}$ individual propane molecules. Why should we use such an absurdly large number? Well, because it is a fact that $6.022 \times {10}^{23}$ individual carbon atoms have a mass of $12 \cdot g$ (or near enuff), and $6.022 \times {10}^{23}$ individual hydrogen atoms have a mass of $1 \cdot g$. These molar masses are listed on the Periodic Table, and there should be a copy beside you now.

Now it follows that a mole of propane SPECIFIES $3 \times 6.022 \times {10}^{23}$ carbon atoms, and $8 \times 6.022 \times {10}^{23}$ hydrogen atoms, and so we multiply thru by the molar masses to give a molar mass for the propane molecule as....

$3 \times 12.011 \cdot g \cdot m o {l}^{-} 1 + 8 \times 1.00794 \cdot g \cdot m o {l}^{-} 1 = 44.10 \cdot g \cdot m o {l}^{-} 1$

That is the background. It is specified that there are $9.36 \times {10}^{24} \cdot \text{hydrogen atoms}$, and thus represents a molar quantity of ..................

$\left(9.36 \times {10}^{24} \cdot \text{hydrogen atoms")/(6.022xx10^23*"hydrogen atoms} \cdot m o {l}^{-} 1\right) = 15.54 \cdot m o l$

And clearly, we can divide this by 8 to give the molar quantity of $1.94 \cdot m o l$ WITH RESPECT TO PROPANE. Do you agree?

And so (finally), there are.....

$1.94 \cdot m o l \times 3 \cdot \text{carbon atoms} \times 6.022 \times {10}^{23} \cdot m o {l}^{-} 1$

$= 3.51 \times {10}^{24} \cdot \text{carbon atoms}$.

And as to the mass of the sample, I take the product...

$\text{Number of moles"xx"molar mass of propane}$

$= 1.94 \cdot m o l \times 44.1 \cdot g \cdot m o {l}^{-} 1 \cong 90 \cdot g$, i.e. I get an answer in grams as required.