8. A 1.871 gram sample of an unknown metallic carbonate is decomposed by heating to form the metallic oxide and 0.656 g of carbon dioxide according to the equation MCO3(s) → MO(s) + CO2 (g) What is the metal (M)?

1 Answer
Meave60
May 6, 2018

The metal is zinc (Zn).

#"ZnCO"_3("s")stackrel(Delta")rarr##"ZnO(s) + CO"_2"#

Explanation:

This question comes from the 2004 U. S. National Chemistry Olympiad Exam Part 1.
https://www.scribd.com/document/280445601/2004-Usnco-Exam-Part-i

There are four answer choices:

A. Ca
B. Mn
C. Ni
D. Zn

Without this information, it would be a daunting task to determine the metal in the compound.

#"MCO"_3("s")stackrel(Delta")rarr##"MO(s) + CO"_2("g")"#

Since you have four possibilities, you can determine which is the correct metal by the process of elimination. The carbonate ion has a #2^(-)# charge. The generic formula for the metal-carbonate compound is given as #"MCO"_3"#, which means that the metal ion has to have a #2^+# charge.

This gives us the following possibilities:

calcium carbonate #("CaCO"_3")#, mangenese(II) carbonate #("MnCO"_3")#, nickel(II) carbonate #("NiCO"_3")#, and zinc carbonate #("ZnCO"_3")#.

You need the molar masses of the possible carbonate compounds. You can calculate them, as would be required on the exam, or you can look them up online since you're not taking the exam. I like to use PubChem. https://pubchem.ncbi.nlm.nih.gov/

#"CaCO"_3:"# #"100.086 g/mol"#

#"MnCO"_3:"# #"114.946 g/mol"#

#"NiCO"_3:"# #"118.701 g/mol"#

#"ZnCO"_3:"# #"125.388 g/mol"#

You also need the molar mass of #"CO"_2"#, which is #"44.09 g/mol"#.

You will need to use stoichiometry to determine the mass of #"CO"_2# that #"1.871 g"# of each metal carbonate can produce.

  1. Determine mol of each metal carbonate by dividing it by its molar mass. Since molar mass is a fraction (g/mol), I prefer to multiply by the inverse of the molar mass (mol/g).

  2. Determine mol of #"CO"_2# produced by multiplying mol metal carbonate by the mol ratio between the metal carbonate and #"CO"_2# from the balanced equation, with #"CO"_2# in the numerator.

  3. Determine mass #"CO"_2# produced by multiplying mol #"CO"_2# by its molar mass.

#1.871color(red)cancel(color(black)("g CaCO"_3))xx(1color(red)cancel(color(black)("mol CaCO"_3)))/(100.086color(red)cancel(color(black)("g CaCO"_3)))xx(1color(red)cancel(color(black)("mol CO"_2)))/(1color(red)cancel(color(black)("mol CaCO"_3)))xx(44.009"g CO"_2)/(1color(red)cancel(color(black)("mol CO"_2)))="0.8227 g CO"_2#

#1.871color(red)cancel(color(black)("g MnCO"_3))xx(1color(red)cancel(color(black)("mol MnCO"_3)))/(114.946color(red)cancel(color(black)("g MnCO"_3)))xx(1color(red)cancel(color(black)("mol CO"_2)))/(1color(red)cancel(color(black)("mol MnCO"_3)))xx(44.009"g CO"_2)/(1color(red)cancel(color(black)("mol CO"_2)))="0.7163 g CO"_2#

#1.871color(red)cancel(color(black)("g NiCO"_3))xx(1color(red)cancel(color(black)("mol NiCO"_3)))/(118.701color(red)cancel(color(black)("g NiCO"_3)))xx(1color(red)cancel(color(black)("mol CO"_2)))/(1color(red)cancel(color(black)("mol NiCO"_3)))xx(44.009"g CO"_2)/(1color(red)cancel(color(black)("mol CO"_2)))="0.6937 g CO"_2#

#1.871color(red)cancel(color(black)("g ZnCO"_3))xx(1color(red)cancel(color(black)("mol ZnCO"_3)))/(125.388color(red)cancel(color(black)("g ZnCO"_3)))xx(1color(red)cancel(color(black)("mol CO"_2)))/(1color(red)cancel(color(black)("mol ZnCO"_3)))xx(44.009"g CO"_2)/(1color(red)cancel(color(black)("mol CO"_2)))=color(red)("0.6567 g CO"_2#

The metal is zinc (Zn).

#"ZnCO"_3("s")stackrel(Delta")rarr##"ZnO(s) + CO"_2"#

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