# Question 6c0de

Sep 18, 2015

Here's what's going on.

#### Explanation:

You know that you have a mixture of barium sulfate, ${\text{BaSO}}_{4}$, and nickel(II) carbonate, ${\text{NiCO}}_{3}$, that has a mass of 2.13 g.

You also know that you manage to separate this mixture and end up with 1.02 g of barium sulfate and 0.81 g of nickel(II) sulfate.

What you need to determine now is what percent of the mixture you managed to recover as barium sulfate, on one hand, and as nickel(II) carbonate, on the other.

To do that, use the mass of each recovered compound and the total mass of the mixture. This will get you

• For barium sulfate

(1.02color(red)(cancel(color(black)("g"))))/(2.13color(red)(cancel(color(black)("g")))) * 100 = color(green)("47.9%")

• For nickel(II) carbonate

(0.81color(red)(cancel(color(black)("g"))))/(2.13color(red)(cancel(color(black)("g")))) * 100 = color(green)("38.0%")

So, you managed to recover 47.9% of the mixture as barium sulfate and 38.0% of the mixture as nickel(II) carbonate.

Assuming that the mixture only contains these two compounds, the total mass you recovered is

${m}_{\text{recovered}} = {m}_{B a S {O}_{4}} + {m}_{N i C {O}_{3}}$

${m}_{\text{recovered" = 1.02 + 0.81 = "1.83 g}}$

This means that you lost a total mass of

${m}_{\text{lost" = m_"mixture" - m_"recovered}}$

${m}_{\text{lost" = 2.13 - 1.83 = "0.30 g}}$

The percent error will thus be

(0.30color(red)(cancel(color(black)("g"))))/(2.13color(red)(cancel(color(black)("g")))) * 100 = color(green)("14.1%")#

The answers are rounded to three sig figs.