Find the Density at STP NaHCO_3

$N a H C O_{3}$

Step 1

To find the mass of $1$ mole of $N a H C O_{3}$ 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 $N a H C O_{3}$ $=$ $(mass of Na)$ + $(mass of H)$ + $(mass of C)$ + $3 (mass of O)$

Step 2

Fill in the atomic masses from the periodic table.

mass of $1$ mole of $N a H C O_{3}$ $=$ $(22.98999977)$ + $(1.00999999)$ + $(12.01000022)$ + $3 (16)$

Step 3

Simplify the result.

Remove parentheses.

mass of $1$ mole of $N a H C O_{3} = 22.98999977 + 1.00999999 + 12.01000022 + 3 (16) g$

Multiply $3$ by $16$ .

mass of $1$ mole of $N a H C O_{3} = 22.98999977 + 1.00999999 + 12.01000022 + 48 g$

Add $22.98999977$ and $1.00999999$ .

mass of $1$ mole of $N a H C O_{3} = 23.99999976 + 12.01000022 + 48 g$

Add $23.99999976$ and $12.01000022$ .

mass of $1$ mole of $N a H C O_{3} = 36.00999999 + 48 g$

Add $36.00999999$ and $48$ .

mass of $1$ mole of $N a H C O_{3} = 84.00999999 g$

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.

$\frac{84.00999999 g}{22.4 L}$

Step 6

Factor $84.00999999$ out of $84.00999999 g$ .

$\frac{84.00999999 (g)}{22.4 L}$

Step 7

Factor $22.4$ out of $22.4 L$ .

$\frac{84.00999999 (g)}{22.4 (L)}$

Step 8

Separate fractions.

$\frac{84.00999999}{22.4} \cdot \frac{g}{L}$

Step 9

Divide $84.00999999$ by $22.4$ .

$3.75044642 \frac{g}{L}$

Step 10

Combine $3.75044642$ and $\frac{g}{L}$ .

$\frac{3.75044642 g}{L}$

Find the Density at STP NaHCO_3

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