1.2 Amount of substance Flashcards Preview

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Flashcards in 1.2 Amount of substance Deck (17)
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1
Q

List the formulae you can use to find the no. of moles.

A

n=m/Mr moles=mass/molecular mass

n=cv moles=concentration (in moldm^-3) x volume (in dm^3)

2
Q

What is the number of particles representing one mole?

A

6.02x10^23

3
Q

Empirical Formula:
“A compound is found to contain 23.3% magnesium, 30.7% sulfur and 46.0% oxygen. What is the empirical formula of this compound?”

A

Mg S O
23.3% 30.7% 46.0%
(Divide by Mr)
23.3/24.3 30.7/32.1 46.0/16.0

=0.959       =0.956     =2.875
    (Divide by smallest value)
(/0.956)      (/0.956)    (/0.956)
=   1                =1             =3
       \                |              /
                   MgSO₃
4
Q

What is the ideal gas equation?

A
pV= nRT
Where:
p = pressure in kPa
V = volume in m^3
n = moles of gas present
R = the gas constant (8.31 JK^-1mol^-1)
T = The gas' temperature in kelvin.
5
Q

Isotopic abundance:
If a sample of boron is composed of 19.9% atoms with an Ar of 10, and 80.1% atoms with an Ar of 11, calculate the atomic weight of boron.

A

(19.9x10) + (80.1x11) / 100

= 10.8 (1dp)

6
Q

How to remember which formulae to use:

A

FOR SOLUTIONS: n=cv
FOR SOLIDS: n=m/Mr
FOR GASES:pV=nRT

7
Q

Calculate the volume of 0.50 moldm^-3 nitric acid required to react completely with 5 g of lead (II) carbonate.

PbCO3 + 2HNO3 –> Pb(NO3)2 + CO2 + H2O

A

n=m/Mr
Moles of PbCO3 = 5/267 = 0.0187
1:2 ratio- moles of HNO3 = 0.0187 x 2 = 0.0375

v=n/c
Volume of HNO3 = 0.0375/0.5 = 0.075dm^3
(0.075x1000=75cm^3 if answer needs to be in cm^3)

8
Q

What’s the formula for percentage yield?

A

actual yield/maximum theoretical yield x 100

9
Q

What is the atom economy for making hydrogen by reacting coal with steam?

C(s) + 2H2O(g) → CO2(g) + 2H2(g)

A

Answer = 8.3%

12 + (2x18) → 44 + (2x2)
(2x2)/(12+36) x 100 = 8.3%

10
Q

What is the formula for percentage uncertainty?

A

(Absolute uncertainty/Value) x100

11
Q

What is the concentration of the solution if 2.5g of calcium carbonate is dissolved in water, and the solution made up to 0.5dm^3?

A

Answer = 0.0500moldm^-3

n=m/Mr = 2.5/40.1+12.0+(3x16.0) = 0.0250mol
c=n/v = 0.250/0.5 = 0.0500moldm-3
12
Q

Calculate the Mr of a substance with a weight of 90g when there are 5 moles of it, and suggest what the substance could be.

A

Answers = 18, water.

Mr = m/n = 90/5 = 18
Water has an Mr of 18.

13
Q

Deduce the formula of the compound lead (IV) oxide.

A

PbO2

14
Q

What is the mass of sulphur in 1 tonne of H2SO4?

A

Answer = 327kg

1 mole of H2SO4 = 98.1g
1000000 grams in a tonne
No. of moles in a tonne = 1000000/98.1 = 10200mol
10200 x 32.1 = 327000g (327kg)

15
Q

25cm^3 of a solution of 0.1moldm^-3 NaOH reacts with 50cm^3 of a solution of hydrochloric acid. What is the molarity of the acid?

A

Answer = 0.05moldm^-3

NaOH + HCl → NaCl + H2O
Moles of NaOH: n=cv = 0.1 x (25/1000) = 0.0025mol
∴ There are 0.0025 moles of HCl
c=n/v = 0.0025/(50/1000) = 0.05moldm^-3

16
Q

How many moles of NaCl are there in 25cm^3 of a 50gdm^-3 solution?

A
n=m/Mr = 50 / 23.0+35.5 = 0.855moldm^-3
n=cv = 0.855 x (25/1000) = 0.0214mol
17
Q

Sodium carbonate exists in hydrated form, Na₂CO₃.xH2O, in the solid state. 3.5g of a sodium carbonate sample was dissolved in water and the volume made up to 250cm^3. 25.0cm^3 of this solution was titrated against 0.1moldm^-3 HCl and 24.5cm^3 of the acid were required. Calculate the value of x given the equation:
​Na₂CO₃ + 2HCl –> 2NaCl + CO2 + H2O

A

x=10

Moles of HCl = 0.0245 x 0.1 = 0.00245mol
∴ moles of Na₂CO₃ = 0.00245/2 = 0.00123mol
Multiply this by 10.
The Mr of Na₂CO₃ = 106.0
∴ Mass of Na₂CO₃ = 106 x 0.01223 = 1.30g
∴ Mass of water in compound = 3.5-1.30=2.2g

Moles of water = m/Mr = 2.2/18 = 0.122mol
0.122/0.01223 = 10
x=10 (Na₂CO₃•10H2O)