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?

Answer 1

#"87.71 g/mol"#

The first step in this process is to calculate how many moles of carbon dioxide were produced by this decomposition reaction using the ideal gas law equation.

This is how the ideal gas law equation appears.

#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

Now, it is crucial that you make sure you convert every unit you are given to the ones that the universal gas constant uses!

In this instance, you must convert the temperature from degrees Celsius to Kelvin, the volume from milliliters to liters, and the pressure from millibars to atm.

You'll get this

#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)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 #"MCO"_3#.

Currently, group 2 carbonates will decompose to produce carbon dioxide and metal oxide, so you can write

#"MCO"_text(3(s]) -> "MO"_text((s]) + "CO"_text(2(g]) uarr#
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 #"CO"_2#, then it must have also consumed #0.001069# moles of metal carbonate.

One mole of metal carbonate will have a mass of, since you know the mass of the sample. You can find the molar mass of the metal carbonate by dividing the mass of the sample by the number of moles that it contained.

#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, #"CO"_3^(2-)#, to find the molar mass of the metal, #"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"#

Consequently, the metal's molar mass is

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

I'll round the result to four significant figures.

The closest match to this value is strontium, #"Sr"#, which has a molar mass of #"87.62 g/mol"#.
Therefore, your metal carbonate was strontium carbonate, #"SrCO"_3#.
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Answer 2

The molar mass of the metal carbonate is approximately 98.2 g/mol.

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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