Topic 2 Revision Flashcards
Average Atomic Mass from % of Isotopes
A sample of Chlorine contains;
25% Cl-37
75% Cl-35
What is the average atomic mass of Chlorine within the sample?
(25x37)+(75x35)
———————— = 35.5
100
Reacting Masses- Limiting Reactants
A sample of 79.8g of iron (|||) oxide is mixed with 9.36g of carbon and heated. A reaction occurs.
2Fe2O3 + 3C -> 4Fe + 3CO2
Show by calculation that iron (|||) oxide is the limiting reactant.
(Ar values: Fe=55.8, O=16.0, C=12.0)
Step 1) Calculate the moles of each reagent
Fe2O3: 79.8/((2x55.8)+(3x16))= 0.50 mol
C= 9.36/12= 0.78 mol
Step 2) Refer to the stoichiometry of the equation
For every 2Fe2O3 which react, 3C are needed
So for 0.5mol of Fe2O3, we need (0.5x3/2)=0.75 mol of C
Step 3) Determine which is in excess
number of moles of C required: 0.75
number of moles of C from step 1: 0.78
So C is in excess by 0.78-0.75= 0.03mol and iron (|||) oxide is limiting.
Reacting Masses
Magnesium burns in oxygen to form magnesium oxide.
2Mg + O2 -> 2MgO
Calculate the mass of oxygen needed to react with 1 mole of magnesium. What is the mass of the magnesium oxide formed?
Step 1) Write the balanced equation
2Mg + O2 -> 2MgO
Step 2) Multiply each formula mass in g by the relevant stoichiometric number in the equation.
2Mg: 2x24.3g= 48.6g
O2: 1x32.0g= 32.0g
2MgO: 2(24.3+16.0g)= 80.6 g
Mass Spectrometry- Number of Carbon Atoms
An unknown compound has a molecular ion peak, M+, with a relative abundance of 54.5% and has an [M+1]+ peak with a relative abundance of 3.6%. How many carbon atoms does the unknown compound contain?
1) Use the formula:
n= (100 / 1.1) * ((abundance of [M+1]+ ion) / (abundance of M+ ion))
2)
n=(100/1.1) * (3.6/54.5)
n=6.0
There are 6 carbon atoms in each molecule
Empirical Formula from Masses
When 1.55g of phosphorus is completely combusted, 3.55g of an oxide of phosphorus is produced. Deduce the empirical formula of this oxide of phosphorus.
(Ar values: O=16.0, P=31.0)
Step 1) Mass of each element
P: 1.55g
O: 3.55g-1.55g= 2.00g
Step 2) Divide by Ar’s
P: 1.55g/31.0=0.05 mol
O: 2.00g/16.0=0.125 mol
Step 3) Divide by the lowest figure
P: 0.05/0.05= 1
O: 0.125/0.05=2.5
Step 4) Obtain the lowest whole number ratio to get empirical formula
=P2O5
Working out the Stoichiometry of a Reaction from Exact Masses in a Reaction
56.2g of silicon, Si, reacts exactly with 284.0g of chlorine, Cl2, to form 340.2 of silicon (IV) chloride, SiCl4. Use this information to calculate the stoichiometry of the reaction.
(Ar values: Cl=35.5, Si=28.1)
Si + Cl2 -> SiCl4
1) Find the number of moles for each substance
Si: 56.2/28.1= 2mol
Cl: 284/(35.5+35.5)= 4mol
Si: 340.2/(28.1+(35.5x4))=2mol
2) Use these ratios as the ratio of stoichiometric numbers in the equation
Si + 2Cl2 -> SiCl4
Percentage Composition by Mass
Calculate the percentage by mass of iron in iron (|||) oxide, Fe2O3.
(Ar values: Fe=55.8, O=16.0)
% mass of iron=
2x55.8 ------------------------ x100= 69.9% (2x55.8) + (3x16.0)
Reacting Masses and Percentage Yield
A sample of aluminium chloride, AlCl3, is made by reacting 18g of aluminium powder with excess chlorine. The mass of aluminium chloride produced is 71.0g. Calculate the percentage yield of aluminium oxide.
(Ar values: Al=27.0, Cl=35.5)
Step 1) Calculate the predicted mass from the stoichiometry of the equation if all the aluminium is converted to aluminium chloride.
2x27g Al produces
2x(27+35.5))g AlCl3
54g Al produces 267g AlCl3
Step 2) Calculate the mass of aluminium chloride formed from the given amount of aluminium using simple proportion
18g of Al produces 267x(18/54)=89.0g AlCl3
Step 3) Calculate the percentage yield
(71.0/89.0)x100=79.8%
Empirical Formula from Percentages
A compound of carbon and hydrogen contains 85.7% carbon and 14.3% hydrogen by mass. Deduce the empirical formula of this hydrocarbon.
(Ar values: C=12.0, H=1.0)
Step 1: Note the % by mass
C: 85.7
H: 14.3
Step 2: Divide by the Ar values
C: 85.7/12.0=7.142
H: 14.3/1.0=14.3
Step 3: Divide by the lowest figure
C: 7.142/7.142=1
H: 14.3/7.142=2
Empirical formula is CH2
Molecular Formula from Empirical Formula and Molar Mass
A compound has the empirical formula CH2Br. Its relative molecular mass is 187.8. Deduce the molecular formula of this compound.
(Ar values: Br=79.9, C=12.0, H=1.0)
Step 1: Find the empirical formula mass
12.0+(2x1.0)+79.9=93.9
Step 2: Divide the relative molecular mass by the empirical formula mass
187.8/93.9= 2
Step 3: Multiply the number of atoms in the empirical formula by the number in step 2
2xCH2Br, so the molecular formula is C2H4Br2.
Writing an Ionic Equation
Step 1: Write down the full balanced equation
Mg(s) + 2HCl(aq) -> MgCl2(aq) + H2(g)
Step 2: Write down all the ions present. Any reactant or product that has a state symbol (s), (l) or (g) is a covalet molecule in solution such as chlorine, Cl2(aq) does not split into ions.
Mg(s) + 2H+(aq) + 2Cl-(aq) -> Mg2+(aq) + 2Cl-(aq) + H2(g)
Step 3: Cancel the ions that appear on both sides of the equation- these are spectator ions.
Mg(s) + 2H+(aq) -> Mg2+(aq) + H2(g)
How to Balance an Ionic Equation
Write the ionic equation for the reaction of aqueous chlorine with aqueous potassium bromide. The products are aqueous bromine and aqueous potassium chloride.
Step 1: The full balanced equation is
Cl2(aq) + 2KBr(aq) -> Br2(aq) + 2KCl(aq)
Step 2: The ions present are
Cl2(aq) + 2K+(aq) + 2Br-(aq) -> Br2(aq) + 2K+(aq) + 2Cl-(aq)
Step 3: Cancel the spectator ions
Cl2(aq) + 2Br-(aq) -> Br2(aq) + 2Cl-(aq)
Step 4: Write the final ionic equation
Cl2(aq) + 2Br-(aq) -> Br2(aq) + 2Cl-(aq)
n = v * c
Calculate the concentration in moldm-3 of sodium hydroxide, NaOH, if 250cm3 of a solution contains 2.0g of sodium hydroxide.
(Mr value: NaOH=40.0)
Step 1: change grams to moles
2.0/40.0= 0.050 mol NaOH
Step 2: Change cm3 to dm3
250cm3= 0.25dm3
Step 3: Calculate concentration
0.050mol/0.25dm3= 0.2moldm-3
Using n=v*c and n=m/Mr
Calculate the mass of anhy
pg 62 worked example 22
Titration Calculation
pg 64 worked example 23
