# Question #7deed

Sep 9, 2016

The block of carbon contains $1.7 \times {10}^{24} \text{atoms}$.

#### Explanation:

You need to determine the density of the carbon. Then use its density and volume to determine the mass of carbon in the block, then determine the number of moles of carbon in the block, and then determine the number of atoms. 1 mole of carbon is equal to $6.022 \times {10}^{23} \text{atoms}$.

Carbon exists as different allotropes; amorphous carbon, graphite, diamond, fullerenes and others. Because graphite is more common, I'm going to use the density of graphite near room temperature, which is $\text{2.2670 g/cm"^3}$. https://en.wikipedia.org/wiki/Carbon; http://education.jlab.org/itselemental/ele006.html

$\text{density"="mass"/"volume}$

$\text{mass"="density"xx"volume}$

Determine the mass of the carbon block.

$\text{mass"="2.2670 g C"/cancel"cm"^3xx15 cancel"cm"^3 "C=34.005 g C}$
(I will round to two significant figures at the end.)

Determine the moles of the carbon in the block.

Divide the mass of the carbon block by the molar mass of carbon. The molar mass of carbon is its atomic mass on the periodic table in g/mol. Carbon's molar mass is $\text{12.0107 g/mol}$.

$\text{34.005 g C"/"12.0107 g/mol"="2.831225 mol C}$

Determine the number of atoms of carbon in the block.

To determine the number of atoms of carbon in the block, multiply the number of moles by $6.022 \times {10}^{23} \text{atoms/mol}$.

$2.831225 \cancel{\text{mol C"xx(6.022xx10^23"atoms C")/cancel"mol C"=1.7 xx10^24 "atoms C}}$ (rounded to two significant figures due to $\text{15 cm"^3}$)