Find the Density at STP H_3BO_3

$H_{3} B O_{3}$

Step 1

To find the mass of $1$ mole of $H_{3} B O_{3}$ 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 $H_{3} B O_{3}$ $=$ $3 (mass of H)$ + $(mass of B)$ + $3 (mass of O)$

Step 2

Fill in the atomic masses from the periodic table.

mass of $1$ mole of $H_{3} B O_{3}$ $=$ $3 (1.00999999)$ + $(10.81000041)$ + $3 (16)$

Step 3

Simplify the result.

Remove parentheses.

mass of $1$ mole of $H_{3} B O_{3} = 3 (1.00999999) + 10.81000041 + 3 (16) g$

Simplify each term.

Multiply $3$ by $1.00999999$ .

mass of $1$ mole of $H_{3} B O_{3} = 3.02999997 + 10.81000041 + 3 (16) g$

Multiply $3$ by $16$ .

mass of $1$ mole of $H_{3} B O_{3} = 3.02999997 + 10.81000041 + 48 g$

Simplify by adding numbers.

Add $3.02999997$ and $10.81000041$ .

mass of $1$ mole of $H_{3} B O_{3} = 13.84000039 + 48 g$

Add $13.84000039$ and $48$ .

mass of $1$ mole of $H_{3} B O_{3} = 61.84000039 g$

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.

$\frac{61.84000039 g}{22.4 L}$

Step 6

Factor $61.84000039$ out of $61.84000039 g$ .

$\frac{61.84000039 (g)}{22.4 L}$

Step 7

Factor $22.4$ out of $22.4 L$ .

$\frac{61.84000039 (g)}{22.4 (L)}$

Step 8

Separate fractions.

$\frac{61.84000039}{22.4} \cdot \frac{g}{L}$

Step 9

Divide $61.84000039$ by $22.4$ .

$2.7607143 \frac{g}{L}$

Step 10

Combine $2.7607143$ and $\frac{g}{L}$ .

$\frac{2.7607143 g}{L}$

Find the Density at STP H_3BO_3

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