A 22.472 g sample of impure magnesium carbonate was heated to complete decomposition according to the equation

MgCO3(s) → MgO(s) + CO2(g)

After the reaction was complete, the solid residue (consisting of MgO and the original impurities) had a mass of 18.42 g. Assuming that only the magnesium carbonate had decomposed, how much magnesium carbonate was present in the original sample? (round to thousandth place)

22.472 g = mass MgCO3 + impurities

-18.42 g = mass MgO + impurities
-----------------
4.05 g = mass CO2 lost.
Convert mass CO2 to mass MgCO3 you started with.
4.05 g CO2 x (molar mass MgCO2/molar mass CO2) = ? g pure MgCO3 initially.
The problem tells you to round to the thousands place; however, your 18.42 is good only to the hundredths place. You may have inadvertently left a zero (or some other number) off the 18.42 g You may need to recalculate the 4.05 number if that's the case.
Then

To find the amount of magnesium carbonate present in the original sample, we can use stoichiometry and compare the masses before and after the reaction.

Let's break down the steps to solve this problem:

1. Calculate the mass of MgO formed:
Mass of MgO = Mass of solid residue - Mass of impurities
Mass of MgO = 18.42 g - mass of impurities (We will calculate the mass of impurities later)

2. Determine the moles of MgO formed:
Moles of MgO = Mass of MgO / Molar mass of MgO
The molar mass of MgO is calculated by adding the atomic masses of Mg and O from the periodic table.

3. Use stoichiometry to find the moles of MgCO3:
According to the balanced equation: 1 mol MgCO3 → 1 mol MgO
So, Moles of MgCO3 = Moles of MgO

4. Calculate the mass of MgCO3:
Mass of MgCO3 = Moles of MgCO3 * Molar mass of MgCO3

Now, let's calculate these step by step:

Step 1: Calculate the mass of MgO formed:
Given: Mass of solid residue = 18.42 g
Assuming only MgCO3 decomposed, the mass of impurities is not considered. So, we use the entire mass of the solid residue.

Mass of MgO = 18.42 g

Step 2: Determine the moles of MgO formed:
Molar mass of MgO = Atomic mass of Mg + Atomic mass of O
= 24.305 g/mol + 16.00 g/mol
= 40.305 g/mol

Moles of MgO = Mass of MgO / Molar mass of MgO
= 18.42 g / 40.305 g/mol

Step 3: Use stoichiometry to find the moles of MgCO3:
Since MgO and MgCO3 have a 1:1 molar ratio according to the balanced equation, the moles of MgO formed will be equal to the moles of MgCO3.

Moles of MgCO3 = Moles of MgO

Step 4: Calculate the mass of MgCO3:
Molar mass of MgCO3 = Atomic mass of Mg + Atomic mass of C + (3 * Atomic mass of O)
= 24.305 g/mol + 12.01 g/mol + (3 * 16.00 g/mol)
= 84.305 g/mol

Mass of MgCO3 = Moles of MgCO3 * Molar mass of MgCO3

Now, you can substitute the values and calculate the answer. Let me know if you need further assistance with the calculations.