# You heat 0.158g of a white, solid carbonate of a Group 2 metal and find that the evolved CO_2 has a pressure of 69.8 mmHg In a 285ml flask at 25°C .What is the molar mass of the metal carbonate?

Dec 3, 2015

$\text{87.71 g/mol}$

#### Explanation:

The first thing you need to do here is use the ideal gas law equation to determine how many moles of carbon dioxide were produced by this decomposition reaction.

The ideal gas law equation looks like this

$\textcolor{b l u e}{P V = n R T} \text{ }$, where

$P$ - the pressure of the gas
$V$ - the volume it occupies
$n$ - the number of moles of gas
$R$ - the universal gas constant, usually given as $0.0821 \left(\text{atm" * "L")/("mol" * "K}\right)$
$T$ - the temperature of the gas

Now, it is of the utmost importance to make sure that you convert all the units given to you to those used by the universal gas constant!

In this case, you need to convert the pressure from mmHg to atm, the volume from milliliters to liters, and the temperature from degrees Celsius to Kelvin.

This will get you

$P V = n R T \implies n = \frac{P V}{R T}$

n = (69.8/760 color(red)(cancel(color(black)("atm"))) * 285 * 10^(-3) color(red)(cancel(color(black)("L"))))/(0.0821 (color(red)(cancel(color(black)("atm"))) * color(red)(cancel(color(black)("L"))))/("mol" * color(red)(cancel(color(black)("K")))) * (273.15 + 25)color(red)(cancel(color(black)("K")))) = "0.001069 moles CO"_2

Next, you need to write a general form balanced chemical equation for the decomposition of this carbonate. Group 2 metal cations have a $2 +$ charge, which means that you can represent the carbonate as ${\text{MCO}}_{3}$.

Now, group 2 carbonates will undergo decomposition to form the metal oxide and carbon dioxide. This means that you can write

${\text{MCO"_text(3(s]) -> "MO"_text((s]) + "CO}}_{\textrm{2 \left(g\right]}} \uparrow$

Notice that you have a $1 : 1$ mole ratio between all the chemical species that take part in the reaction. This means that if the reaction produced $0.001069$ moles of ${\text{CO}}_{2}$, then it must have also consumed $0.001069$ moles of metal carbonate.

Since you know the mass of the metal carbonate sample, you can find its molar mass by dividing this value by the number of moles that it contained. Simply put, one mole of metal carbonate will have a mass of

1 color(red)(cancel(color(black)("moles MCO"_3))) * "0.158 g"/(0.001069 color(red)(cancel(color(black)("moles MCO"_3)))) = "147.8 g"

Finally, use the molar mass of the carbonate anion, ${\text{CO}}_{3}^{2 -}$, to find the molar mass of the metal, $\text{M}$.

overbrace(1 xx "12.0107 g/mol")^(color(blue)("one atom of carbon")) + overbrace(3 xx "15.9994 g/mol")^(color(red)("three atoms of oxygen")) = "60.09 g/mol"

Therefore, the molar mass of the metal is

"147.8 g/mol" - "60.09 g/mol" = color(green)("87.71 g/mol")

I'll leave the answer rounded to four sig figs.

The closest match to this value is strontium, $\text{Sr}$, which has a molar mass of $\text{87.62 g/mol}$.

Therefore, your metal carbonate was strontium carbonate, ${\text{SrCO}}_{3}$.