A 0.60 g sample consisting of only CaC2O4 & MgC2O4 is heated at 5000C, converting the two salts of CaCO3 and MgCO3 . The sample then weighs 0.465 g. If the sample had been heated to 9000C, where the products are CaO & MgO, what would the mixtures of oxides have weighed?
a) 0.12 g b) 0.21 g c) 0.252 g d) 0.3 g
There are several ways to do this and this method I'm showing you may be the longest but it's the first one I came up with. So here goes. This is a two equation problem and you solve the two equations simultaneously. Here are the steps.
1. Set up equations to calculate g CaC2O4 and g MgC2O4.
2. Then convert g CaC2O4 to g CaO
3. Convert g MgC2O4 to g MgO
4. Add g CaO + g MgO to find total. The correct answer is in the choices on your list.
Here are the equations.
Let X = mass CaC2O4
and Y = mass MgC2O4, then
-------------------
eqn 1 is X + Y = 0.60g
For eqn 2 you want to chemically convert X (which is CaC2O4) to CaCO3 and Y(which is MgC2O4) to MgCO3 because you know what the carbonates weigh.
MM stands for molar mass.
(MM CaCO3/MM CaC2O4)X + (MM MgCO3/MM MgC2O4)Y = 0.465g
Solve those two equations to find X and Y.
Then Xg CaCO3 can be converted to g CaO by (MM CaO/MM CaCO3)X = ?
Yg MgCO3 can be converted to g MgO by
(MM MgO/MM MgCO3)Y = ?
Then add g CaO to g MgO to find total.
You should obtain an answer VERY close to one of the choices.
Post your work if you get stuck.
0.252
To find the weight of the mixture of oxides when heated to 900°C, we need to calculate the weight of CaO and MgO separately and then add them up.
Let's start with calculating the weight of CaO:
1. Calculate the weight of CaC2O4 in the original sample by subtracting the weight of the final sample (0.465 g) from the initial sample weight (0.60 g):
Weight of CaC2O4 = 0.60 g - 0.465 g = 0.135 g
2. Calculate the molar mass of CaC2O4 by adding up the atomic masses of each element:
Molar mass of CaC2O4 = (40.08 g/mol) + 2*(12.01 g/mol) + 4*(16.00 g/mol) = 128.12 g/mol
3. Use the molar mass to calculate the number of moles of CaC2O4:
Moles of CaC2O4 = Weight of CaC2O4 / Molar mass of CaC2O4 = 0.135 g / 128.12 g/mol
4. Since CaO is formed from CaCO3 by the loss of CO2, we can calculate the molecular weight of CO2:
Molecular weight of CO2 = (12.01 g/mol) + 2*(16.00 g/mol) = 44.01 g/mol
5. Calculate the weight of CaO formed from CaC2O4 by the loss of CO2:
Weight of CaO = Moles of CaC2O4 * (Molar mass of CO2 - Molar mass of CaO)
= Moles of CaC2O4 * (44.01 g/mol - 56.08 g/mol)
= Moles of CaC2O4 * (-12.07 g/mol)
Now, let's calculate the weight of MgO:
1. Calculate the weight of MgC2O4 in the original sample by subtracting the weight of the final sample (0.465 g) from the initial sample weight (0.60 g):
Weight of MgC2O4 = 0.60 g - 0.465 g = 0.135 g
2. Calculate the molar mass of MgC2O4 by adding up the atomic masses of each element:
Molar mass of MgC2O4 = (24.31 g/mol) + 2*(12.01 g/mol) + 4*(16.00 g/mol) = 148.32 g/mol
3. Use the molar mass to calculate the number of moles of MgC2O4:
Moles of MgC2O4 = Weight of MgC2O4 / Molar mass of MgC2O4 = 0.135 g / 148.32 g/mol
4. Since MgO is formed from MgCO3 by the loss of CO2, we can calculate the molecular weight of CO2:
Molecular weight of CO2 = (12.01 g/mol) + 2*(16.00 g/mol) = 44.01 g/mol
5. Calculate the weight of MgO formed from MgC2O4 by the loss of CO2:
Weight of MgO = Moles of MgC2O4 * (Molar mass of CO2 - Molar mass of MgO)
= Moles of MgC2O4 * (44.01 g/mol - 40.31 g/mol)
= Moles of MgC2O4 * (3.70 g/mol)
Finally, add the weights of CaO and MgO:
Weight of mixture of oxides = Weight of CaO + Weight of MgO
Now, you can calculate the weight of the mixture of oxides by plugging in the calculated values for each salt and performing the addition. The final answer should be one of the options given (0.12 g, 0.21 g, 0.252 g, or 0.3 g).