Question

 

Exercise 1: Stoichiometry and a Precipitation Reaction

Data Table 1. Stoichiometry Values.

Initial: CaCl2•2H2O (g)	1.0
Initial: CaCl2•2H2O (moles)	.0068
Initial: CaCl2 (moles)	.0067
Initial: Na2CO3 (moles)	.0068
Initial: Na2CO3 (g)	.7209
Theoretical: CaCO3 (g)	.68
Mass of Filter paper (g)	1.0
Mass of Filter Paper + CaCO3 (g)	1.62
Actual: CaCO3 (g)	.62
% Yield:	91.17%

Questions

A perfect percent yield would be 100%. Based on your results, describe your degree of accuracy and suggest possible sources of error.

The yield percentage of my experiment was a little over 91%. This means that approximately 9% was error, because maybe some of the precipitate didn’t filter through or some was still left in the beaker.

What impact would adding twice as much Na₂CO₃ than required for stoichiometric quantities have on the quantity of product produced?

There will be no impact because CaCl₂ is the limiting reagent. The product weight would still be the same.

Determine the quantity (g) of pure CaCl₂ in 7.5 g of CaCl₂•9H₂O.

Mass of CaCl₂•9H₂O is 273.12 g/mol

Mass of CaCl₂ is 110.98 g/mol

Moles of CaCl₂•9H₂O is 7.5g/273.12 g/mol= .0275 moles CaCl₂•9H₂O

So, .0275 moles• 110.98 g/mol= 3.05 g of CaCl₂

Determine the quantity (g) of pure MgSO₄ in 2.4 g of MgSO₄•7H₂O.

Mass of MgSO₄•7H₂O is 246.47 g/mol

Mass of MgSO₄ is 120.37

Moles of MgSO₄•7H₂O is 2.4g MgSO₄•7H₂O•120.37g MgSO₄/ 246.47g MgSO₄•7H₂O= 1.17 g MgSO₄

Conservation of mass was discussed in the background. Describe how conservation of mass (actual, not theoretical) could be checked in the experiment performed.

Conservation of the mass can be obtained by calculating the initial mass of the CaCl₂ and with that you can find the mass of the initial Ca and then find the mass of the final product. The mole ratio should already be known, so we can find the mass of Ca in the product and compare it with the initial mass of the Ca. which was used in the starting of the experiment.

Answer 1

first lets calculate the CaCl2 2H2O, it has a molar mass of 147 g/gmol

moles = mass / molar mass = 1 gram / 147 = 0.0068 moles

you have 0.0068 moles of Na2CO3 , it has amolar mass of 106 g/gmol

moless = mass / molar mass, mass = moles * molar mass

mass = 106 * 0.0068 = 0.7208 grams

The CaCl₂ will react with 1 mole of Na2CO3 in a 1:1 reaction, so every mole of CaCl₂ will produce 1 mole of CaCO₃ for each mole available so we will have 0.0068 moles of CaCO₃ , this compound has a molar mass of 100 g/gmol

mass of CaCO3 = moles * molar mass = 0.0068 * 100 = 0.68 grams

you got 0.62 grams of real yield so %yield will be

= 0.62 / 0.68 = 0.91 or 91.17% , your values are correct.

You didnt get a 100% yield because some of the reactant was lost in the beaker, the reaction needed more time to complete. etc

Adding to much of one reactant, Na2CO3 in this case will not make any difference because the production of CaCO3 depens on 2 reactants, the Na2CO3 will be limited by the other reactant.

The CaCl2*9H2O has a molar mass of 273 g/gmol , the CaCl2 has a molar mass of 111 g/gmol, lets get the percent of CaCl2 in CaCl2 9H2O

111 / 273 = 0.4

so 40% of the weight comes from the CaCl2 so, calculate the grams of CaCl2, it will be the 40% of 7.5,

7.5 grams * 0.4 = 3 grams of CaCl₂

The case for MgSO4 is similar

MgSO4 7H2O has a molar mass of 246.47 g/gmol

MgSO4 has a molar mass of 120.37

fraction of MgSO4 = 120.37 / 246.47 = 0.488

2.4 * 0.488 = 1.1712 grams of MgSO4

conservation of mass might be checked by comparing the initial ammounts of CaCl2, and the final concentrations for the products CaCo3 and the CaCl2 unreacted, if you add the moles of CaCo3 produced and CaCl2 unreacted this must be equal to the moles of CaCl2 available from the beginning