Chemistry Examples

$C H_{3} C O O H$

Step 1

To find the mass of $1$ mole of $C H_{3} C O O H$ look up the atomic mass of each element and multiply it by the number of atoms contained in each element in the molecule.

mass of $1$ mole of $C H_{3} C O O H$ $=$ $(mass of C)$ + $3 (mass of H)$ + $(mass of C)$ + $(mass of O)$ + $(mass of O)$ + $(mass of H)$

Step 2

Fill in the atomic masses from the periodic table.

mass of $1$ mole of $C H_{3} C O O H$ $=$ $(12.01000022)$ + $3 (1.00999999)$ + $(12.01000022)$ + $(16)$ + $(16)$ + $(1.00999999)$

Step 3

Simplify the result.

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Step 3.1

Remove parentheses.

mass of $1$ mole of $C H_{3} C O O H = (12.01000022) + 3 (1.00999999) + 12.01000022 + 16 + 16 + 1.00999999 g$

Step 3.2

Multiply $3$ by $1.00999999$ .

mass of $1$ mole of $C H_{3} C O O H = 12.01000022 + 3.02999997 + 12.01000022 + 16 + 16 + 1.00999999 g$

Step 3.3

Add $12.01000022$ and $3.02999997$ .

mass of $1$ mole of $C H_{3} C O O H = 15.0400002 + 12.01000022 + 16 + 16 + 1.00999999 g$

Step 3.4

Add $15.0400002$ and $12.01000022$ .

mass of $1$ mole of $C H_{3} C O O H = 27.05000042 + 16 + 16 + 1.00999999 g$

Step 3.5

Add $27.05000042$ and $16$ .

mass of $1$ mole of $C H_{3} C O O H = 43.05000042 + 16 + 1.00999999 g$

Step 3.6

Add $43.05000042$ and $16$ .

mass of $1$ mole of $C H_{3} C O O H = 59.05000042 + 1.00999999 g$

Step 3.7

Add $59.05000042$ and $1.00999999$ .

mass of $1$ mole of $C H_{3} C O O H = 60.06000041 g$

Step 4

One mole of any gas occupies $22.4 L$ at STP (standard temperature and pressure).

$V = 22.4 L$

Step 5

Density is mass over volume.

$\frac{60.06000041 g}{22.4 L}$

Step 6

Factor $60.06000041$ out of $60.06000041 g$ .

$\frac{60.06000041 (g)}{22.4 L}$

Step 7

Factor $22.4$ out of $22.4 L$ .

$\frac{60.06000041 (g)}{22.4 (L)}$

Step 8

Separate fractions.

$\frac{60.06000041}{22.4} \cdot \frac{g}{L}$

Step 9

Divide $60.06000041$ by $22.4$ .

$2.68125001 \frac{g}{L}$

Step 10

Combine $2.68125001$ and $\frac{g}{L}$ .

$\frac{2.68125001 g}{L}$