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?

Question

Mateo Aguilar · Accepted Answer

#"148 g/mol"#

The idea behind this is that in order to calculate how many moles of carbon dioxide were produced by the reaction, you must apply the ideal gas law equation.

Once you know how many moles of carbon dioxide there are in the reaction, you can use the balanced chemical equation to go back and find out how many moles of the metal carbonate were broken down.

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#

The ideal gas law equation appears like this, as you are aware.

#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 situation, you will need to convert temperature from degrees Celsius to Kelvin, pressure from milliliters to liters, and volume from milliliters to atm.

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"#.

Accordingly, the mass of one mole of the metal carbonate will be

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

As a result, your metal carbonate's molar mass equals

#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.

Avery Adams · Answer

undefined The molar mass of the Group 2 metal carbonate is approximately 110.8 g/mol.