A sample of sandstone consists of silica, SiO2, and calcite, CaCO3. When the sandstone is heated, calcium carbonate, CaCO3, decomposes into calcium oxide, CaO, and carbon dioxide.
CaCO3(s) → CaO(s) + CO2(g)
What is the percentage of silica in the sandstone if 19.2 mg of the rock yields 4.21 mg of carbon dioxide?
Convert 4.21 mg CO2 to moles. 1 mole CO2 = 1 mole CaCO3, then convert moles CaCO3 to grams CaCO3.
%CaCO3 = (g CaCO3/0.0192 g sample)*100 = ??
% SiO2 = 100%-%CaCO3.
Post your work if you have any trouble.
By the way, there is a MUCH easier way to do this but it isn't taught anymore since everyone has gone to SI units.
The easy way to convert CO2 to CaCO3 is
4.21 mg CO2 x (molar mass CaCO3/molar mass CO2) = 4.21 x (100/44) = ?? mg CaCO3. That molar mass CaCO3/molar mass CO2 is called the "gravimetric factor" but it isn't in any of the modern texts.
Thanks so much. These examples really help me out.
Well, let's do some math and unravel this mystery, shall we?
First, we need to determine the amount of calcium carbonate (CaCO3) that decomposed to produce 4.21 mg of carbon dioxide (CO2). Since the molar mass of CaCO3 is 100.09 g/mol and the molar mass of CO2 is 44.01 g/mol, we can set up a proportion:
(4.21 mg CO2 / 44.01 g/mol CO2) = (x mg CaCO3 / 100.09 g/mol CaCO3)
Solving for x, we find that x = (4.21 mg CO2 / 44.01 g/mol CO2) * (100.09 g/mol CaCO3) ≈ 4.02 mg CaCO3.
Now that we know the amount of CaCO3 that reacted, we can subtract this from the initial weight of the sandstone sample (19.2 mg) to find the weight of SiO2:
Weight of SiO2 = 19.2 mg - 4.02 mg = 15.18 mg.
To determine the percentage of SiO2 in the sandstone, we can use the formula:
Percentage of SiO2 = (Weight of SiO2 / Initial weight of sandstone) * 100%.
Percentage of SiO2 = (15.18 mg / 19.2 mg) * 100% ≈ 79.2%.
So, the percentage of silica (SiO2) in the sandstone is approximately 79.2%. Just remember, sandstone might be stubborn, but it can still be measured and calculated with a touch of mathematical comedy!
To find the percentage of silica in the sandstone, we need to determine the amount of silica present in the rock sample.
First, we calculate the amount of carbon dioxide produced from the decomposition of the calcium carbonate, using the given data:
Mass of CO2 = 4.21 mg
Next, we need to determine the amount of calcium carbonate that was decomposed. From the balanced chemical equation, we can see that 1 mole of calcium carbonate (CaCO3) produces 1 mole of carbon dioxide (CO2). The molar mass of CaCO3 is 100.09 g/mol.
Molar mass of CaCO3 = 100.09 g/mol
Using the molar mass, we can convert the mass of CO2 to moles of CaCO3:
moles of CaCO3 = mass of CO2 / molar mass of CaCO3
Now, we can calculate the moles of CaCO3:
moles of CaCO3 = 4.21 mg / 100.09 g/mol = 0.0421 mmol
Since the reaction is 1:1 between CaCO3 and SiO2, the moles of CaCO3 also represent the moles of SiO2.
Therefore, the moles of SiO2 = 0.0421 mmol.
To determine the mass of SiO2, we need to use the molar mass of SiO2. Its molar mass is 60.08 g/mol.
Molar mass of SiO2 = 60.08 g/mol
Mass of SiO2 = moles of SiO2 × molar mass of SiO2
Mass of SiO2 = 0.0421 mmol × 60.08 g/mol = 2.53 mg
Finally, we can calculate the percentage of silica in the sandstone:
Percentage of SiO2 = (Mass of SiO2 / Mass of sandstone) × 100
Since the mass of the sandstone is given as 19.2 mg:
Percentage of SiO2 = (2.53 mg / 19.2 mg) × 100
Percentage of SiO2 ≈ 13.18%
The percentage of silica in the sandstone sample is approximately 13.18%.
For my explanation I will use "What is the percentage of silica in the sandstone if 18.0 GRAMS of the rock yields 4.4 GRAMS of carbon dioxide?" as the question. Note that you will have to convert mg to g.
First, balance the equation and write it down. This equation is already balanced, so 1 CaCO3--------> 1 CaO + 1 CO2.
Next, ask yourself "What information do I need in order to solve the problem?" We need 2 of the following 3 values:
1) mass of sandstone, we know is 18g
2) mass of silica in the sandstone
3) mass of CaCO3 originally in the sandstone
We are only given the mass of the sandstone, therefore, there must be a way to determine one of the other two values from the information we were presented. In our case, we will deduce the amount of CaCO3 originally present using the amount of CO2 produced.
In any balanced equation, the coefficients of the substances directly relate to the molar ratio between any substances involved. That is, we can say "One mole of CaCO3, upon decomposition, will produce exactly one mole of CO2," because each has a coefficient of 1 in the balanced equation. These statements can be made in "reverse" as well. If we get one mole of CO2 produced, we know that there was one mole of CaCO3 present originally.
Once you realize that a problem is going to be dealing with moles, grams, etc., write down each substance involved, along with its molecular weight and how much of the substance you have.
CaCO3----------mol. wt. 100-------------______ grams
CaO--------------mol. wt. 56--------------______ grams
CO2--------------mol. wt. 44----------------- 4.4 grams
SiO2-------------mol. wt. 60.1--------------_____ grams
Sandstone-------not known------------------18.0 grams
If we know how many moles of CO2 were produced, then we know how many moles of CaCO3 were present initially.
We are given CO2 in grams, to convert to moles use the following:
...............[1 mol CO2 ]
4.4g CO2 [-----------------] = 0.1 moles of CO2
...............[ 44g CO2 ]
Since the coefficients are the same, we know that this translates to 0.1 moles of CaCO3 initially present. To convert to grams:
...........................[ 100g CaCO3 ]
0.1 moles CaCO3 [---------------------] = 10g CaCO3
...........................[1 mol CaCO3 ]
If you start with moles, then put moles on the bottom and molecular weight on top. The molecular weight of CaCO3 is 100g/mol. The moles cancel out leaving grams. If you start with grams, put grams on the bottom and moles on top.
Put these values in the table as you go.
CaCO3----------mol. wt. 100-------------__10__ grams
CaO--------------mol. wt. 56--------------______ grams
CO2--------------mol. wt. 44----------------- 4.4 grams
SiO2-------------mol. wt. 60.1--------------_____ grams
Sandstone-------not known------------------18.0 grams
We now know 2 of the 3 original values that were needed. Mass of sandstone = 18g, and mass of CaCO3 = 10g.
Since the sandstone is composed of only two substances, CaCO3 and SiO2, figuring one percentage will automatically give us the other. Percent composition =
Mass part
------------------------ =
Mass total
10g CaCO3
--------------------------- = 56% CaCO3, leaving 44% silica.
18g sandstone
Now work through the problem step by step with your values. Don't forget to convert the mg to grams.