Quantitative Part 1 Flashcards
Using Avogadros constant (6.02x10^23 mol^-1), what’s the number is atoms in 1 mil of dinitrogen tetroxide (N2O4)?
N2O4 has 6 atoms
Therefore - (6.02x10^23) x 6 = 3.614 x 10^24 ✔️
The equation for the COMPLETE combustion of ethane is:
2C2H6(g) + 7O2(g) —> 4CO2(g) + 6hH2O(l)
What volume of Oxygen measures at RTP would be needed to completely burn 0.1 mol of ethane ?
(0.1/2) x 7 = n
n= 0.35
Sub n into n=Vol/24
Vol= n x 24
0.35 x 24 = 8.4dm^3 ✔️
1.12g of iron reacts with oxygen to form 1.60g of an oxide of iron. Use relative atomic masses:
Fe = 56, O = 16
What’s the formula of this oxide of iron?
n = mass/mr
n(Fe) = 1.12/56 = 0.02 n(O) = 48/16 = 0.03
Therefore: Fe2O3 ✔️
In an experiment, 1.226g of Pottasium chlorate(V), (KClO3), was heated. A mass of 0.320g of oxygen gas was collected.
2KclO3(s) —> 2KCl(s) + 3O2(g)
Use the molar mass of KClO3 = 122.6gmol^-1 and relative atomic mass O = 16.0
Find the percentage yield.
% yield = (act/theor) x 100
Mol for 2KClO3 =
1.226/122.6 = 0.01
Mol for 3O2 = (0.01/2) x 3 = 0.48
(0.320/0.480) x 100 = 66.7% ✔️
Oxygen can be prepared using several different reactions. Whoch reaction has the highest atom economy by mass?
A- NaNO3 —> NaNO2 + 0.5O2
B- H2O2 —> H2O + 0.5O2
C- Cl2 + H2O —> HCl + 0.5O2
D- PbO2 —> PbO + 0.5O2
B because needs to have the highest % therefore the rproduct needs to have lowest Mr.
Oxygen gas (O2) can be converted into ozone (O3) by passing it through an electric discharge, 3O2(g) —> 2O3(g).
In an experiment, a volume of 300cm^3 of Oxygen was used but only 10% of the oxygen was converted into ozone. All volumes were measured at the same RTP.
Why was the Total volume of gas present at the end of the experiment (in cm^3) ?
•First, find 10% of 300 and take it off - 300 x 0.9 = 270cm^3
- Find the ratio - 3:2 (3O2 —> 2O3)
- Find the gas volume that wasn’t turned into ozone : 300 - 270 = 30cm^3
30 ? 30/3 = 10 (x2) = 20 = ?
3 : 2
•Lastly find total gas present at end of experiment
20 + 270 = 290cm^3 ✔️
Chloric(V) acid has the following percentage composition by mass:
H, 1.20% ; Cl, 42.0% ; O, 56.8%
Using this information, calculate the empirical formula of chloric(V) acid.
•First, % over Mr H = 1.20/1 = 1.20mol Cl = 42.0/35.5 = 1.18mol O = 56.8/16 = 3.55mol •Divide by the smallest and simplify 1.20/1.18 = H = 1 1.18/1.18 = Cl = 1 3.55/1.18 = O = 3
Empirical formula = HClO3
Oxygen can be prepared using several different reactions. Whoch reaction has the highest atom economy by mass?
A- NaNO3 —> NaNO2 + 0.5O2
B- H2O2 —> H2O + 0.5O2
C- Cl2 + H2O —> HCl + 0.5O2
D- PbO2 —> PbO + 0.5O2
B because needs to have the highest % therefore the rproduct needs to have lowest Mr.
Oxygen gas (O2) can be converted into ozone (O3) by passing it through an electric discharge, 3O2(g) —> 2O3(g).
In an experiment, a volume of 300cm^3 of Oxygen was used but only 10% of the oxygen was converted into ozone. All volumes were measured at the same RTP.
Why was the Total volume of gas present at the end of the experiment (in cm^3) ?
•First, find 10% of 300 and take it off - 300 x 0.9 = 270cm^3
- Find the ratio - 3:2 (3O2 —> 2O3)
- Find the gas volume that wasn’t turned into ozone : 300 - 270 = 30cm^3
30 ? 30/3 = 10 (x2) = 20 = ?
3 : 2
•Lastly find total gas present at end of experiment
20 + 270 = 290cm^3 ✔️
Chloric(V) acid has the following percentage composition by mass:
H, 1.20% ; Cl, 42.0% ; O, 56.8%
Using this information, calculate the empirical formula of chloric(V) acid.
•First, % over Mr H = 1.20/1 = 1.20mol Cl = 42.0/35.5 = 1.18mol O = 56.8/16 = 3.55mol •Divide by the smallest and simplify 1.20/1.18 = H = 1 1.18/1.18 = Cl = 1 3.55/1.18 = O = 3
Empirical formula = HClO3
Oxygen can be prepared using several different reactions. Whoch reaction has the highest atom economy by mass?
A- NaNO3 —> NaNO2 + 0.5O2
B- H2O2 —> H2O + 0.5O2
C- Cl2 + H2O —> HCl + 0.5O2
D- PbO2 —> PbO + 0.5O2
B because needs to have the highest % therefore the rproduct needs to have lowest Mr.
Oxygen gas (O2) can be converted into ozone (O3) by passing it through an electric discharge, 3O2(g) —> 2O3(g).
In an experiment, a volume of 300cm^3 of Oxygen was used but only 10% of the oxygen was converted into ozone. All volumes were measured at the same RTP.
Why was the Total volume of gas present at the end of the experiment (in cm^3) ?
•First, find 10% of 300 and take it off - 300 x 0.9 = 270cm^3
- Find the ratio - 3:2 (3O2 —> 2O3)
- Find the gas volume that wasn’t turned into ozone : 300 - 270 = 30cm^3
30 ? 30/3 = 10 (x2) = 20 = ?
3 : 2
•Lastly find total gas present at end of experiment
20 + 270 = 290cm^3 ✔️
Chloric(V) acid has the following percentage composition by mass:
H, 1.20% ; Cl, 42.0% ; O, 56.8%
Using this information, calculate the empirical formula of chloric(V) acid.
•First, % over Mr H = 1.20/1 = 1.20mol Cl = 42.0/35.5 = 1.18mol O = 56.8/16 = 3.55mol •Divide by the smallest and simplify 1.20/1.18 = H = 1 1.18/1.18 = Cl = 1 3.55/1.18 = O = 3
Empirical formula = HClO3
What’s the ideal gas equation ?
Pv=nRT
0.11g of pure barium was added to 100cm^3 of water
Ba(s) + 2H2O(l) —> Ba(OH)2(aq) + H2
- A) Show that 8x10^-4 mol of Ba was added to the water.
- B) Calculate the Volume of H2 has produced at RTP (in cm^3)
A) n = mass/Mr
0.11/137.3 = 8.01x10^-4 ✔️
B) n = Vol/24 therefore, Vol = 8.01x10^-4 x 24
= 0.01 cm^3
MgSO4•xH2O is heated to remove the water. 1.57g of water was removed, leaving 1.51g of anhydrous MgSO4.
- A) calculate the amount (in mol) of anhydrous MgSO4 formed.
- B) calculate the amount (in mol) of H2O removed.
A) mol = mass/mr
1.51/120.4 = 0.0125 mol
B) mol = mass/mr
1.57/18 = 0.872 mol
70.0μg of iodine (70x10^-6g) is the guidelines to take daily in order to help the thyroid gland function properly. A company decides to mix iodide with table salt (KI) as a source of the ions.
Calculate the total mass of KI needed (in μg) to supply the thyroid with the correct amount of ions.
(Give your answer in 3.s.f)
•First calculate mol of iodine (using g)
Mol = mass/mr
70x10^-6/126.9 = 551.6x10^-9
•find mr of both K and I combined and find total mass using mol
Mr(KI) = 166
551.6x10^-9 x 166 = 91.5x10^-6g
Convert back into μg = 91.5μg ✔️
A student heats 12.41g of hydrated Na2S2O3•5H2O, to remove the water of crystallisation. A white powdery called anhydrous sodium thiosulphate forms.
- A) what does anhydrous mean?
- B) calculate the expected mass of anhydrous sodium thiosulphate that forms.
A) without water
B) find the mol of the solution (mol = mass/mr)
12.41/248.2 = 0.05 mol
Find mass of sodium thiosulphate (mass = mr x mol)
158.2 x 0.05 = 7.81 ✔️
A sample of ammonia gas occupied a volume of 1.53 x 10^-2 at 37 degrees and pressure of 100KPa
Calculate the amount (in mol) of ammonia in this sample.
Pv = nRT
n = Pv/RT
Convert KPa into Pa (x1000)
100,000 x (1.53x10^-2) /
What’s meant by the term standard solution?
A solution of known concentration ✔️✔️
