# 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 69.8mmHg in a 285ml flask at 25°C. What is the molar mass of the metal carbonate?

##### 1 Answer

#### Explanation:

The idea here is that you need to use the ideal gas law equation to determine how many **moles** of carbon dioxide were produced by the reaction.

Once you know the number of moles of carbon dioxide, you can backtrack using the *balanced chemical equation* for this reaction to determine how many **moles** of the metal carbonate underwent decomposition.

So, a metal located in group 2 of the periodic table will form

#["M"]^(2+)["CO"_3]^(2-) implies "MCO"_3#

Metal carbonates undergo thermal decomposition to form a *metal oxide*, in your case *carbon dioxide*,

#"MCO"_text(3(s]) stackrel(color(red)("heat")color(white)(xx))(->) "MO"_text((s]) + "CO"_text(2(g]) uarr#

As you know, the ideal gas law equation looks like this

#color(blue)(PV = nRT)" "# , where

*number of moles* of gas

**always** expressed in *Kelvin*

Before plugging in your values into this equation, make sure that the units given to you **match** those used in the expression of

In your case, you will need to convert pressure from *mmHg* to *atm*, volume from *milliliters* to *liters*, and of course temperature from *degrees Celsius* to *Kelvin*.

Rearrange the above equation to solve for

#PV = nRT implies n = (PV)/(RT)#

#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 + 25.0)color(red)(cancel(color(black)("K"))))#

#n = "0.001069 moles CO"_2#

So, your reaction produced

#0.001069 color(red)(cancel(color(black)("moles CO"_2))) * "1 mole MCO"_3/(1color(red)(cancel(color(black)("mole CO"_2)))) = "0.001069 moles MCO"_3#

As you know, **molar mass** is defined as the mass occupied by **one mole** of a given substance. In your case, the sample that contained

This means that **one mole** of the metal carbonate will have a mass of

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

Therefore, the *molar mass* of your metal carbonate is equal to

#M_M = color(green)("148 g/mol") -># rounded to three sig figs

The closest match to this value is *strontium carbonate*,