# Question ff2b8

Apr 15, 2015

There are 3.8 × 10^23" C atoms" in 10 g of $\text{CH₄}$ and 8.2 × 10^23" C atoms" in 20 g of ${\text{C"_3"H}}_{8}$.

$\text{C}$ atoms in ${\text{CH}}_{4}$

You must convert

$\text{grams of CH"_4 → "moles of CH"_4 → "molecules of CH"_4 → "atoms of C}$

The molar mass of $\text{CH"_4}$ is 16.04 g/mol.

You use this to convert grams to moles.

10 cancel("g CH"_4) × (1 cancel("mol CH₄"))/(16.04 cancel("g CH₄")) × (6.022 × 10^23 cancel("molecules CH₄"))/(1 cancel("mol CH₄")) × "1 atom C"/(1 cancel("molecule CH₄")) = 3.8 × 10^23"atoms C"

C atoms in ${\text{C"_3"H}}_{8}$

The molar mass of ${\text{C"_3"H}}_{8}$ is 44.10 g/mol.

Can you show there are 8.2 × 10^23" C" atoms in 20 g of ${\text{C"_3"H}}_{8}$?

Apr 15, 2015

Hi, the first thing that you want to do is to change the grams of each compound into moles.

To do this, you divide the grams by their corresponding molar mass.

Next you multiply the mole values by Avogadro's number, 6.02 × 10^23.

This will tell you the number of molecules of each compound.

Finally you multiply the number of molecules of each compound by the number of carbon atoms in one molecule of each compound.

${\text{CH}}_{4}$:

(10 cancel("g"))/(16 cancel("g")"/mol") = "0.625 mol"

0.625 cancel("mol") × (6.02 × 10^23"molecules")/(1 cancel("mol")) = 3.76 × 10^ 23"molecules"

3.76 × 10^ 23 cancel("molecules") × "1 C atom"/(1 cancel("molecule")) =

3.76 × 10^23 " C atoms in 10 g of "CH"_4.

${\text{C"_3"H}}_{8}$:

(20 cancel("g"))/ (44 cancel("g")"/mol") = "0.455 mol"

0.455 cancel("mol") × (6.02 × 10^23"molecules")/(1 cancel("mol")) = 2.74 × 10^23 "molecules"

2.74 × 10^23 cancel("molecules") × "3 C atoms"/(1 cancel("molecule")) =

8.2 × 10^23" C atoms in 20 g of "C_3"H"_8#

Apr 16, 2015

Dear Peggy, Please note that in the explanation on how to solve the problem. At the very end, the answer is multiplied by 3 since there are 3 Carbons in C3H8.

Apr 16, 2015

Thank Jay P. Good luck on your test!