# A student had a sample of "BaCl"_2 * 2"H"_2"O" and an inert material. He heated a sample until the mass did not change...? (Problem below)

## From this data, calculate the percentage of BaCl2•2H2O in the original sample. • Mass of crucible + cover = 26.39 g • Mass of crucible + cover + unknown sample = 28.20 g • Mass of crucible + cover + residue = 28.03 g

Feb 3, 2017

63.68%

#### Explanation:

The idea here is that you can use the mass of the residue to figure out the mass of barium chloride present in the sample.

When you heat the barium chloride dihydrate and the inert material, the water of crystallization evaporates and leaves behind anhydrous barium chloride, ${\text{BaCl}}_{2}$, and the inert material.

Now, the mass of the barium chloride dihydrate + inert material mixture will be equal to

${m}_{\text{mixture" = m_"crucible + cover + mixture" - m_"crucible + cover}}$

In your case, this is equal to

${m}_{\text{mixture" = "28.20 g" - "26.39 g}}$

${m}_{\text{mixture" = "1.81 g}}$

After you heat the mixture, the water of crystallization is driven off

${m}_{\text{water" = m_"crucible + cover + mixture" - m_"crucible + cover + residue}}$

This is equal to

${m}_{\text{water" = "28.20 g" - "28.03 g}}$

${m}_{\text{water" = "0.17 g}}$

Now, you know that one mole of barium chloride dihydrate contains

• one mole of anhydrous barium chloride, $1 \times {\text{BaCl}}_{2}$
• two moles of water of crystallization, $2 \times \text{H"_2"O}$

Use the molar mass of water to figure out the number of moles that evaporate, i.e. the number of moles of water that were present in the hydrate before heating the sample

0.17 color(red)(cancel(color(black)("g"))) * ("1 mole H"_2"O")/(18.015color(red)(cancel(color(black)("g")))) = "0.009437 moles H"_2"O"

This means that the number of moles of barium dihydrate present in the sample is equal to

0.009437 color(red)(cancel(color(black)("moles H"_2"O"))) * ("1 mole BaCl"_2 * 2"H"_2"O")/(2color(red)(cancel(color(black)("moles H"_2"O"))))

$= \text{0.0047185 moles BaCl"_2 * 2"H"_2"O}$

Use the molar mass of barium chloride dihydrate to find the number of moles of hydrate present in the sample

0.0047185 color(red)(cancel(color(black)("moles BaCl"_2 * 2"H"_2"O"))) * "244.26 g"/(1color(red)(cancel(color(black)("mole BaCl"_2 * 2"H"_2"O"))))

$= \text{1.15254 g BaCl"_2 * 2"H"_2"O}$

You already know that the initial mixture had a mass of $\text{1.81 g}$, which means that the percent composition of barium chloride dihydrate in the sample was

"% BaCl"_2 * 2"H"_2"O" = (1.15254 color(red)(cancel(color(black)("g"))))/(1.81color(red)(cancel(color(black)("g")))) * 100% = color(darkgreen)(ul(color(black)(63.68%)))

The answer is rounded to four sig figs.