Find the Density at STP C_14H_14N_3O_3SNa

$C_{14} H_{14} N_{3} O_{3} S N a$

Step 1

To find the mass of $1$ mole of $C_{14} H_{14} N_{3} O_{3} S N a$ 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 $C_{14} H_{14} N_{3} O_{3} S N a$ $=$ $14 (mass of C)$ + $14 (mass of H)$ + $3 (mass of N)$ + $3 (mass of O)$ + $(mass of S)$ + $(mass of Na)$

Step 2

Fill in the atomic masses from the periodic table.

mass of $1$ mole of $C_{14} H_{14} N_{3} O_{3} S N a$ $=$ $14 (12.01000022)$ + $14 (1.00999999)$ + $3 (14.01000022)$ + $3 (16)$ + $(32.06999969)$ + $(22.98999977)$

Step 3

Simplify the result.

Remove parentheses.

mass of $1$ mole of $C_{14} H_{14} N_{3} O_{3} S N a = 14 (12.01000022) + 14 (1.00999999) + 3 (14.01000022) + 3 (16) + 32.06999969 + 22.98999977 g$

Simplify each term.

Multiply $14$ by $12.01000022$ .

mass of $1$ mole of $C_{14} H_{14} N_{3} O_{3} S N a = 168.1400032 + 14 (1.00999999) + 3 (14.01000022) + 3 (16) + 32.06999969 + 22.98999977 g$

Multiply $14$ by $1.00999999$ .

mass of $1$ mole of $C_{14} H_{14} N_{3} O_{3} S N a = 168.1400032 + 14.13999986 + 3 (14.01000022) + 3 (16) + 32.06999969 + 22.98999977 g$

Multiply $3$ by $14.01000022$ .

mass of $1$ mole of $C_{14} H_{14} N_{3} O_{3} S N a = 168.1400032 + 14.13999986 + 42.03000068 + 3 (16) + 32.06999969 + 22.98999977 g$

Multiply $3$ by $16$ .

mass of $1$ mole of $C_{14} H_{14} N_{3} O_{3} S N a = 168.1400032 + 14.13999986 + 42.03000068 + 48 + 32.06999969 + 22.98999977 g$

Simplify by adding numbers.

Add $168.1400032$ and $14.13999986$ .

mass of $1$ mole of $C_{14} H_{14} N_{3} O_{3} S N a = 182.28000307 + 42.03000068 + 48 + 32.06999969 + 22.98999977 g$

Add $182.28000307$ and $42.03000068$ .

mass of $1$ mole of $C_{14} H_{14} N_{3} O_{3} S N a = 224.31000375 + 48 + 32.06999969 + 22.98999977 g$

Add $224.31000375$ and $48$ .

mass of $1$ mole of $C_{14} H_{14} N_{3} O_{3} S N a = 272.31000375 + 32.06999969 + 22.98999977 g$

Add $272.31000375$ and $32.06999969$ .

mass of $1$ mole of $C_{14} H_{14} N_{3} O_{3} S N a = 304.38000345 + 22.98999977 g$

Add $304.38000345$ and $22.98999977$ .

mass of $1$ mole of $C_{14} H_{14} N_{3} O_{3} S N a = 327.37000322 g$

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.

$\frac{327.37000322 g}{22.4 L}$

Step 6

Factor $327.37000322$ out of $327.37000322 g$ .

$\frac{327.37000322 (g)}{22.4 L}$

Step 7

Factor $22.4$ out of $22.4 L$ .

$\frac{327.37000322 (g)}{22.4 (L)}$

Step 8

Separate fractions.

$\frac{327.37000322}{22.4} \cdot \frac{g}{L}$

Step 9

Divide $327.37000322$ by $22.4$ .

$14.61473228 \frac{g}{L}$

Step 10

Combine $14.61473228$ and $\frac{g}{L}$ .

$\frac{14.61473228 g}{L}$

Find the Density at STP C_14H_14N_3O_3SNa

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