Amount Of Substance

Question 1

One mole contains 6.02 x 10^23 atoms - what is this known as ?

Answer 1

Avagadros constant

Question 2

What is the equation for number of particles?

Answer 2

Number of moles x Avagadros constant

Question 3

What is the equation for number of moles?

Answer 3

Mass of substance / RFM

concentration (mol dm-3) x volume (dm3)

Question 4

Ideal gas equation ?

Answer 4

pV = nRT

Question 5

What do each variable stand for in pV = nRT ?

Answer 5

P - pressure (Pa)
V - volume (m^3)
n - number of moles
R - 8.31 J K-1mol-1 (always given)
T - temperature (K)

Question 6

At a temperature of 60 degrees and a pressure of 250 kPa, a gas occupied a volume of 1100cm^3 and had a mass of 1.6g.

find its relative molecular mass

the gas constant is 8.31 J K-1mol-1

Answer 6

p = 250 x 1000 = 250 x10^3 Pa

V = 1100 x (1x10^-6) = 1.1 x 10^3 m^3

T = 60 +273 = 333 K

n = pV/RT

(250 x 10^3) x (1.1 x 10^3) / 8.21 x 333 = 0.0994 moles

Mr of gas = mass / number of moles = 1.6 g / 0.0994 = 16.1

Question 7

Balance the equation

C2H6 + O2 –> CO2 + H2O

Answer 7

C2H6 + 3.5 O2 –> 2 CO2 + 3H2O

Question 8

ionic equations only show ______ particles

Answer 8

reacting

example

HNO3 + NaOH –> NaNO3 + H2O

break up into ions

H+ + NO3- + Na+ + OH- –> Na+ + NO3- +H2O

cancel out

H+ + OH- –> H2O

Question 9

Balanced equations can be used to work out masses

calculate mass of iron oxide produced if 27.9g pf iron is burnt in air.

4Fe + 3O2 –> 2Fe2O3

Answer 9

Mr of Fe = 55.8
mass/ RFM
27.9/55.8 = 0.5 moles
4 moles of Fe gives 2 moles of Fe2O3 so 0.5 moles of Fe would give 0.25 moles of Fe2O3
Mr of Fe2O3 =(2 x 55.8) + ( 3 x 16.0) = 159.6
moles x Mr = 0.25 x 159.6 = 39.9g

Question 10

How to make a standard solution for titration ?

Answer 10

1. Measure out solid on a balance, in a weighing boat
2. Add into a beaker and wash weighing boat with distilled water, and add distilled water and stir with a glass rod to make sure the solid dissolves
3. Pour into a volumetric flask, using a funnel
4. Rinse beaker to make sure nothing is left behind
5. Fill to 250 ml / to line with more distilled water until the bottom of the meniscus touches the line
6. Invert to mix

Question 11

titration method?

Answer 11

1. Do a rough titration to get an idea of the end point. Add the acid to alkali using a burette, giving the flask a regular swirl
2. Take an initial reading o how much acid is in the burette. \run the acid in to within 2cm3 of the end point. When you get to this stage add it in dropwise and keep swirling until the colour changes
3. Work out the amount of acid used to neutralise the alkali. Final - initial, this volume is known as the titre
4. Repeat the titration a few times until you have concordant results
5. Use the results from each concordant result to calculate the mean volume of acid used.

Question 12

indicators for titration

Methyl orange turns _____

phenolphthalein turns _____

Answer 12

red in acid –> yellow in alkali

colourless in acid –> pink in alkali

Question 13

In a titration experiment, 25cm3 of 0.5 mol dm-3 HCl neutralised 35cm3 of NaOH solution

Calculate the concentration of the sodium hydroxide solution in mol dm-3

Answer 13

first write. balanced equation and write what you have

HCl + NaOH –> NaCl + H2O
25cm3 35 cm3
0.5 ??
moldm-3

mol of HCl 0.5 x 25/1000 = 0.0125 moles

from the equation we know that 1 mole of HCl neutralises 1 mole of NaOH
so 0.0125 moles of HCl. must neutralise 0.0125 moles of NaOH

conc of NaOH = moles / vol (dm3) = 0.0125 / (35/1000) = 0.357 mol dm-3

Question 14

Define empirical formula

Answer 14

simplest whole number ratio of atoms of each element in a compound

Question 15

Define molecular formula

Answer 15

actual number of atoms of each element in a compound

Question 16

A molecule with Mr = 166 has the empirical formula of C4H3O2

find molecular formula

Answer 16

(4 x 12) + (3 x 1) + (2 x 16) = 48 + 3 + 32 = 83

166/83

= 2 empirical units in the molecule

molecular formula must be empirical x 2 so C8H6O4

Question 17

A compound is found to have percentage composition

56.5% potassium
8.7% carbon
34.8% oxygen

find its empirical formula

Answer 17

in 100g there would be
mass/ Mr
56.5/39.1 = 1.445 mol of K
8.7/12 = 0.725 mol of C
34.8/16 = 2.175 mol of O

divide each number by the smallest
K = 2.00
C = 1.00
O = 3.01

ratio is 2 : 1 : 3

K2CO3

Question 18

Percentage yield = _______________ / _____________ x ___

Answer 18

actual yield / theoretical yield x 100

Question 19

0.475 g of CH3Br reacts with excess NaOH in the following reaction

CH3Br + NaOH –> CH3OH + NaBr
0.153 g of CH3OH is produced

What is the percentage yield?

Answer 19

number of moles of CH3Br = mass/ Mr = 0.475 / (12 + ( 3 x 1) + 79.9 ) +0.00501 mol

from the equation, moles of CH3OH is 1:1, so 0.00501 moles of CH3OH should form
Mr of CH3OH = 32
Theoretical yield = 0.00501 x 32 = 0.160 g

so

0.153 / 0.160 x 100 = 95.6%

Question 20

% atom economy = ???

Answer 20

molecular mass of desired product / sum of molecular masses of all reactants x 100

Question 21

Ethanol ( C2H12OH) can be produced by fermenting glucose (C6H12O6)

C6H12O6 –> 2C2H5OH + 2CO2

calculate atom economy

Answer 21

Always make sure the equation is balanced

2 x ((2 x 12 ) + ( 5 x 1) + ( 16 + 1 )) / ( 6 x 12) + ( 12 x 1 ) + (6 x 16) x 100 =

92/180 x 100 = 51.1%

companies want a high atom economy - less waste, less expensive, more sustainable