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CH 101/Keys 3

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CH 101
  Exercises 1  
  Exercises 2  
  Exercises 3  
  Exercises 4  
  Exercises 5  
  Exercises 6  
  Exercises 7  
  Exercises 8  
  Exercises 9  
  Exercises 10  
  Exercises 11  
  Exercises 12  
  Exercises 13  
  Exercises 14  
  Exercises 15  




Contents

Some combined problems

Exercise 1

To a solution of a salt of a lanthanide metal M 1 mmole Br is added in the form of bromide. This is a small amount compared to the metal in solution. A precipitate is formed and weighed after careful separation and drying. Its mass is 132.4mg. It can be assumed that
  1. the precipitate has the chemical formula is MBr3
  2. that all bromine added precipitated
What was the element in solution?
Tricky.. We cannot really write out the full balanced equation, because it does not tell what the dissolved salt is. However we do know that all the bromine comes out of solution. This means that the MBr3 precipitate must be 1/3 of a mmole otherwise the bromines do not add up.
That means that the molar mass of the MBr3 precipitate is 3*132.3 = 397.2. [g/mol]
From there we can get the atomic mass of M by subtracting three bromines: 397.2 - 3*80 =157.2
Inspection of the periodic table then tells us this must be Gadolinium.
Of course, we are assuming that there was only one lanthanide element in solution!! If there was a mixture of them and they all came out of solution the atomic mass we find reflects an average over them and we cannot tell what we have without doing a lot more work....


Exercise 2

An aliquot of naphthalene C10H8 is burnt in a bomb pressurized to 25 Atm with pure oxygen. The water formed is collected and weighed. Its weight is 1.324 g. What was the weight of the aliquot of naphthalene, assuming the combustion was complete?
We need to start with a balanced combustion reaction equation. As it involves a combustion of an organic compound we know the products are H2O and CO2. The bit about 25 Atm is nice but irrelevant for the calculation.
C10H8 +12O2 ==> 4H2O and 10CO2
We also need the molar mass for water = 16+2=18 [g/mol]
This means that the combustion gave 1.324 / 18 [g].[mol/g] = 73.6 mmol of water
We need to divide that by 4 to find the number of mmols of naphthalene of the aliquot (1 mole of it gives 4 mole of water!)==> 18.3 mmol of naphthalene
finally we need the molar mass of naphthalene = 10*12+8*1=128 [g/mol] = 128 [mg/mmol]
multiplyling gives us what we want: 128*18.3 [mg/mmol][mmol]= 2345 mg = 2.345 g

Exercise 3

Consider the reaction

CaC2 + H2O ==> Ca(OH)2 + C2H2
All the ethyn C2H2 gas is collected and burnt completely with pure oxygen, in which reaction 4.32 g of water are formed.
How much calcium carbide (by weight) was used assuming that an excess of water was used to react with it?
The tricky part is that we are dealing with two reactions here and neither is balanced yet
equation 1: CaC2 + 2H2O ==> Ca(OH)2 + C2H2
equation 2: C2H2 + 2.5 O2 ==> H2O + 2CO2
Another tricky part is that we are dealing with water in both reactions, but only the water in the second really matters
Using the molar mass of 18 for water and the second reaction we see that we had 4.32/18 [g][mol/g] = 0.24 moles of water
That means that we also had 0.24 moles of ethyn (see equation 2) and that in turn tells us we had 0.24 moles of carbide (see equation 1)! (All the coefficients are 1)
The molar mass of carbide is 40+2*12= 64 [g/mol]
That gives us 0.24*64 [mol][g/mol] = 15.32 gram of calcium carbide.

Exercise 4

Consider the reaction

CaCO3(s) ==> CaO(s) + CO2(g)
The pressure is 1 Atm and temperature 25oC.
What is the volume of carbon dioxide gas evolved when 16kg of calcium oxide is formed?
The balanced equation tells us that for each mole of CaO one mole of carbon dioxide is set free
We need the molar mass of CaO: 40+16= 56 [g/mol] = 56 [kg/kmol]
Dividing we find 16/56 [kg][kmol/kg] = 0.2847 kmol = 284.7 mol
At the conditions given 1 mol of gas equals 22.4 liter
Multiplying we find the volume of the liberated gas = 22.4*284.7 = 6,400 liter.

Exercise 5

A flat piece of titanium with a surface area of 1cm2 is oxidized in oxygen. A 350nm thin layer is formed of anatase TiO2. The density of anatase is 4.3 [g/cm3].

How many titanium atoms does the thin layer contain?

We can find the volume of the layer: V=surface*height, but we must make sure to convert the nanometers to centimeters: 350 nm= 350 10-9m = 350 10-7cm = 3.50 10-5cm
Thus the volume V= 1*3.50 10-5[cm2][cm]=3.50 10-5[cm3]
Now we can calculate its mass, using the density:
mass = 3.50 10-5*4.3 [cm3][g/cm3] = 15.05 10-5 [g] = 150.5 [μg]
The molar mass of anatase = 48+2*16=80 [g/mol]= 80 [μg/μmol]
Dividing gives: 150.5/80 = [μg][μmol/μg]=1.88[μmol]
Using NAvo, we find 1.88 10-6. 6.2022 1023 = 11.6 1017 atoms of Ti

Exercise 6

The compound NH4V3O8 is prepared in three steps:

  1. N2 + H2 ==> NH3
  2. NH3 + V2O5 + H2O ==> NH4VO3
  3. NH4VO3 + HCl ==> NH4V3O8 + NH4Cl + H2O
  1. How many grams of N2 are needed to prduce 610 g of NH4V3O8, assuming all other reactants are present in abundance
  2. How much water is consumed/produced in the (total) synthesis.
First balance the equations!
  1. N2 + 3H2 ==> 2NH3
  2. 2NH3 + V2O5 + H2O ==> 2NH4VO3
  3. 3NH4VO3 + 2HCl ==> NH4V3O8 + 2NH4Cl + H2O
Then convert the grams of the product into moles, because otherwise the balanced equations are useless.
Molar mass NH4V3O8: =14+4*1+3*50.9+8*16 =298.7 [g/mol]
610 g /298.7 [g/mol]= 2.04 [mol] NH4V3O8
Now we have to combine the ratios of the three steps
  1. 1:3::2
  2. 2:1:1::2
  3. 3:2::1:2:1

Look at the NH4VO3, it has a coefficient 3 in the bottom, but a 2 in the middle expression. We can make them equal by multiplying #2 by 3 and #3 by 2:

  1. 1:3::2
  2. 2:1:1::2 (*3)
  3. 3:2::1:2:1 (*2)

We get

  1. 1:3::2
  2. 6:3:3::6
  3. 6:4::2:4:2

This couples equations #2 and #3 as one series of ratios.

Now look at NH3. It now has a 6 in #2, but a 2 in #1. Multiply #1 with 3:
  1. 3:9::6
  2. 6:3:3::6
  3. 6:4::2:4:2

Now all three equations are coupled ratios. Now look at the final product NH4V3O8 on the bottom line and the initial reactant N2 on the first.

  1. 3:9::6
  2. 6:3:3::6
  3. 6:4::2:4:2
This tells us that for every two moles of NH4V3O8 we have to start with three moles of N2. This gives us the conversion factor [3/2] [mol N2.]/[mol NH4V3O8]
We actually produce a little more: 2.04 mole. Using the conversion factor we see it will require 2.04* [3/2]= 3.06 [mol] of N2.
We can also look at the water. In step two water is consumed, in step three water is produced
  1. 3:9::6
  2. 6:3:3::6
  3. 6:4::2:4:2
For every 2 mol of NH4V3O8, 2 mol water is produced in the last step, but 3 mol is consumed in the middle step. That means that in total 1 mol of water is consumed per two moles of product.
For 2.04 mol of product we therefore lose 1.02 mole of water.
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