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
Dec 28, 2015

Answer:

#"148 g/mol"#

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 #2+# cations, #"M"^(2+)#. As you know, the carbonate anions carry a #2-# charge, so you can represent the metal carbonate as

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

Metal carbonates undergo thermal decomposition to form a metal oxide, in your case #"MO"#, and carbon dioxide, #"CO"_2#

#"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

#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("atm" * "L")/("mol" * "K")#
#T# - the temperature of the 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 #R#.

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 #n#

#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# moles of carbon dioxide. Since the metal carbonate decomposes in a #1:1# mole ratio with carbon dioxide, you can say that the reaction used up

#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 #0.001069# moles of metal carbonate has a mass of #"0.158 g"#.

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 #"147.6 g/mol"#, the molar mass of strontium carbonate, #"SrCO"_3#, so you can say that this is a very good result.