Amount of a substance

write 0.0000278 in standard form

2.78 × 10^-5

when doing x// how many sig figs should you give

the least amount in the data

how many s.f should you give normally

3s.f

what is relative atomic mass

the average mass of one atom of an element on a scale chosen so that the mass of the carbon-12 isotope is exactly 12.0000

Ar= average mass of one atom of an element/ 1/12 mass of one atom of carbon 12

what is relative molecular mass

the average mass of one molecule of a compound on a scale chosen so that the mass of the carbon 12 isotopes is exactly 12.0000

Mr = average mass of one molecule of the compound / 1/12 mass of one atom of carbon 12

how many decimal places do you give to Ar and Mr

1 decimal place

how can you work out moles

moles= mass(in grams)/ Mr

what is the mass of 5 moles of chlorine

5 × 71= 355

how many molecules are there in 42.5g on ammonia

Ar= N 14 H 1

Mr= 14+3= 17

42.5/ 17 × 6.02 × 10²³

= 1.51 × 10²⁴

how do you find how many particles in a certain mass of a substances

number of particles= mass g/ molar mass M

x 6.022 × 10²³

a sample of cyclohexane contains 3.011 × 10²⁴ atoms of carbon what is the mass of this sample

C6H12

70g

what is the empirical formula

simplest ratio of atoms of each elemnt in a substance e.g ratio of 1:2

what is a molecular formula

indicates the number of atoms of each element in one molecule

how do you find the molecular formula is you are given the formula mass and the empirical formula

1. divide mr by the formual mass of empirical formula

2. whatever number you get multiply the empirical formula by that

if the empirical formula is CH2 and the Mr 42 find the moleuclar formula

12+2=14

42/14=3

3x CH2= C3H6

a compound is found to contain by mass iron 72.4% and oxygen 27.6%

72.4/ 56= 1.29 27.6/16= 1.73

divide by the smallest valye

1: 1.34

muliply by 3 to get whole numbers

3:4

Fe3O4

0.25g of hydrogen reacts with oxygen to produce 4.25g of hydrogen peroxide mr =34

1. calculate the mass of oxygen and set up your masses of each element

4.25-0.25= 4

2. divide each mass or percentage by the Ar of the element (not Mr)

0.25/1=0.25 4/16= 0.25

3. 1:1 ratio HO empiricla formula

4. calculate the formula mass of the empirical formula

17

5. divide the Mr

34/17=2

H2O2

how do you find the mass of anhydrous

mass after heating- mass of crucible

mass of water

mass before- mass after

what is the difference between anhydrous and hydrous

anhydrous without water

hydrated with water (trapped within crystals ) g.XH2O

calculate the % of water in hydrated magnesium sulphate MgSO₄ . 7H₂O salt crystals

1. calculate the relative formula mass

24+32+(4×16)+(7×18)= 246

2. find the relative mass of water

7× 18= 126

126/246 ×100= 51.2%

6.25g of blue hydrate copper sulphate CuSO₄ . xH₂O was gently heated in a crucible until the remaining mass was 4g find x

6.25-4=2.25g mass of water driven off

moles of copper sulfate and moles of water

4/16=0.25 2.25/18= 0.125

divide by smallest value

1:5 x=5

what mass of iron is produced when 32kg of iron oxide is heated with CO

balanced equation

Fe₂O₃ + 3 CO = 2Fe + 3CO₂

moles of iron oxide 32000/159.6= 200.5 moles

1 iron oxide forms 2 mol iron

2× 200.5= 401.10

401× 55.8= 22400g

what mass of oxygen is needed to convert 102g of ammonia into nitrogen

4NH₃ + 3O₂ = 2N₂ + 6H₂O

102/17=6 moles

4:3

6× 3/4= 4.5 moles

4.5 × 32= 144g

when 5g of crystals of hydrated tin chloride SnCl₂.xH₂O = SnCl₂ + xH₂O are heated 4.2g of anhydrous tin chloride are formed. calculate x

SnCl₂.xH₂O = SnCl₂ + xH₂O

moles of SnCl₂ 4.2/189.7= 0.02214

moles of SnCl₂.xH₂O is the same

5/ 0.02214= 225.8

225.8 -189.7= mr of water= 36.1

36.1/18=2

x=2

what is the limitng reagent

the reactant that isn’t in excess- we don’t have any more than we need

propanone reacts with oxygen C₃H₈ + 5O₂ = 3CO₂ + 4H₂O

how many moles of product are formed when 1 mole of c3h8 is mixed with 8 moles of o2

1:5:3:4

oxygen is in excess as we have 8 moles but only use 5

propanone is the limiting reagent

7 moles of product

in the manufacture of titanium what mass can be made when 1kg of titanium chloride reacts with 0.1kg of magnesium

TiCl₄ + 2Mg = Ti + 2MgCl₂

moles at the start

1000/189.9= 5.266 moles

100/24.3= 4.115 moles

5.266 moles needs 10.53 moles to react so titanium chloride is in excess and magneisum is the limitng reageny

2.058 moles of ticl4 will react

1:1

2.058× 47.9= 98.6g

why may the yield not be 100%

• different side reactions occur

• impure reactions

• solid- lost in transfer

• reversible reaction will not go to competion

how do you find percentage yield

mass of product obtained/ maximum theoretical mass of product x 100

calculate the maximum theoretical mass that can be made from one tonne of iron oxide

only 650kg were made calculate the percentage yield

1000000/159.6= 6266 moles

2×6266=12530

12530× 55.8= 699kg

650/699× 100 = 93%

what is atom economy

a measure of what proportion of the products of a reaction are the desired product and how much is waste. the higher the atom economy the less waste that is produced

how can you find atom economy

mr/mass of desired / mass/mr of all products x 100

what do you use to work out atom economy

THE BIG NUMBERS

calculate the atom economy to make calcium oxide from calcium carbonate

CaCO₃ = CaO + CO₂

56%

what is the ideal gas equation and what are the units

PV= nRT

P= pressure in Pa ( 1 atmosphere = 100KPa)

n= number of moles

R= gas constant (8.31)

T = temperature (K)
V= volume m³

how do you go from degrees to kelvin

add 273

how do you go from cm3 to m3

divide by a million

calculate the pressure exerted by 0.1 moles of an ideal gas at 50 degrees with a volume of 1500cm3

P= nRT/V

0.1 × 8.31x (50+273 ) / (1500/1×10^6) = 179000 Pa

what is 200 degrees in kelvin

473 kelvin

an ideal gas occupies a volume of 2.75 dm3 at 290 k and 8.7×10^4 Pa at what temperature will it occupy 3.95 dm3 at 1.01 × 10^5

P= 8.7 × 10⁴

V= 2.75 × 10^-3

R= 8.31

T= 290

n= x

n= PV/RT

n= 0.099277

P2= 1.01 × 10⁵

V2= 3.95×10^-3

R= 8.31

T= x

N= 0.099

T= PV/nR

= 484K

what does or doesn’t effect the volume of a gas and what does this mean

depends on temperature, pressure and number of moles

what the gas is does not effect its volume

under the same temperature and pressure 100cm3 of one gas contains the same number of moles of any other gas

what volume of oxygen reacts with 100cm3 of but-1-ene

C4H8 +6O2 = 4CO2 + 4H2O

1:6 so 600cm³

1dm³ of but-1-ene is reacted with 10dm³ of oxygen, what volume of oxygen remains at the end

C4H8 +6O2 = 4CO2 + 4H2O

4 dm3 left over

if 4dm3 of hydrogen sulfide is burned in 10dm3 of oxygen what is the final volume of the mixture

2H₂S + 3O₂ = 2H₂O + 2SO₂

2:3

4/2 × 3 =6 10-6=4 so 4dm3 of o2 is unreacted

4 +4+4 = 12 dm3 total at end

how do you find concentration and give all units

concentration= moles/ volume

volume in dm3

concentration moldm-3

what is a monoprotic acid and give an example

contain one H+ ion per unit

HCl

with NaOH they react in the ratio 1:1

what is a diprotic acid and give an example

contains 2 h+ ions per unit

H2SO4

with NaOH they react in the ratio 1:2

what is a triprotic acid and give an example

contains three H+ ions per unit

H3PO4

with NaOH they react in the ratio 1:3

how do you find concentration in g dm-3

do the normal concentration and x by Mr

25cm³ of 0.02 mol dm-3 sulphurinc acid neutralises 18.6cm³ of sodium hydroxide solution

H₂SO₄ + 2NaOH = Na₂SO₄ 2H₂O

Finf the concetration of the sodium hydroxide solution in moldm-3

conc x vol = 0.02 × 0.025 = 0.0005

1:2

2× 0.0005= 0.001

mol/vol = 0.001/0.0186= 0.0538 moldm-3

crystals of citric acid contain water of crystallisation (C₆H₈O_7. nH₂O) Citirc acid is a triprotic acid, 1.52g of the citric acid was made up to 250cm3 solution. 25cm3 portions of this solution required 21.8 cm3 of 0.1 moldm-3 of sodium hydroxide for neutralisation. find n

moles of NaOH= 0.1 × 0.0218= 0.00218

moles of citric acid in the 25cm³ pipete using ratio of the reaction

acid to alakli ratio 1:3

0.00218/3= 0.000727

total solution is ten times bigger than the pippette

0.000727 × 10= 0.00727

mr = mass/moles

1.52/ 0.00727= 209.2

209.2- 192.1= 17.1

n=1

a solution of a metal carbonate M₂CO₃ was prepared dissolving 7.46g on the anhydrous solid in water to give 1000cm³ of solution. 25cm³ of this solution reacted with 27cm³ of 0.1moldm^-3 hydrochloric acid. calculate the relative formula mass of M₂CO₃ and hence the relative atomic mass of the metal M

M₂CO₃ +2HCl = 2MCl + CO₂ +H₂O

HCl- 0.027 × 0.1= 0.0027

divide by 2= 0.00135

25 to 1000 x by 40

0.00135 × 40= 0.054

7.46/0.054= 138.15

c=12 3O=48

138.2-12-48= 78.1 /2 39.1

this is potassium

what is denisty

mass (g)/ volume (cm³)

a 100cm³ solution of glucose needs to be made it has a density of 0.9gcm^-3 what is the mass of glucose that needs to be weighed out? what are the moles of glucose used?

100× 0.9= 90g

90/ (72+12+96)

0.5 moles

what is a back titration

done to analyse a base that does not react quickly with an acid, the base is treated with an excess of acid and then left over acid is titrated. you can then work back to find out about the o.g base

limestone is mainly calcium carbonate. a student wanted to find what percentage of some limestone was calcium carbonate, a 1g sample of limestone is allowed to react with 100 cm3 and 0.2 mol/dm3 HCl. the excess acid required 24.8 cm3 of 0.1 mol/dm3 of NaOH solution in a back titration. calculate the percentage of calcium carbonate in the limestone

CaCO3 + 2HCl = CaCl2 + H2O + CO2 HCl + NaOH = NaCl + H2O

1g 100cm3 24.8cm3

p 0.2mol/dm3 0.1moldm3

24.8/1000 × 0.1 = 2.48 × 10^-3

1:1 ratio

0.1 × 0.2= 0.02 - (2.48 × 10^-3)= 0.01752

1:2 0.01752/2 = 8.76 × 10^-3 moles of calcium carbonate

8.76 × 10^-3 x (40.1 + 12+48)= 0.87678

0.87678/1 × 100 = 87.6%

an impure sample of barium hydroxide of mass 1.6524g was allowed to react with 100cm³ of 0.2mol/dm3 of HCl. when the excess acid was titrated against 0.228 mol/dm3 sodium hydroxide in a back titration 10.9 cm³ of sodium hydroxide was required. what is the percentage purity of barium hydroxide

how many protons are there in 6g of nitrogen gas

1.8 × 10²⁴

^{each nitrogen is diatomic so 14 protons}

compound P is converted into compound r by a 2 stage synthesis via compund q

P to Q = 50%
Q to R = 30% what is the overall yield of r in this

15%

write the half equation for the conversion of NO₂^- in an acidic solution into NO

NO₂^- + 2H⁺ + e- = NO + H₂O

write a half equation for the conversion of I- Into I₂

2I- = I₂ + 2e-

write an overall ionic equation for the reaction of NO₂^- in an acidic solution with I-

2NO₂^- + 4H⁺ +2I^- = 2I₂ + NO + H₂O

suggest how a student doing an experiment could check all the water has been removed

heat to a constant mass

describe a method to mae a standard solution using NaCl and water

1. using a top-pan balance accurate to 0.01g weigh a clean, dry weighing bottle and lid. add the required mass to the bottle and reweigh

2. transfer the crystals to a 100cm3 beaker weigh by difference (to calculate the actual mass of crystal added

3. add around 60cm3 more of distilled water to the beaker and stir with a glass rod until the crystals dissolve

4. transfer using a funnel the contents to a 100cm3 volumetric flask

5. complete 3 washings of the beaker to make sure all of the solutions is added

6. add more to the volumetric flask until it is just below the mark then add drop by drop until the miniscus is one the line

7. invert the volumetric flask 20 times to ensure the solution is mixed thoroughly

explain a titration of HCl and NaOH

1. fill the burette with hydrochloric acid

2. pipette 10cm3 of the sodium hydroxide solution into a 100cm3 conical flask

3. add 3 drops of phenolphthalein indicator to the flask pink to colourless

4. record the intial burette reading to 0.05cm3 accuracy

5. titrate the sodium hydorxide solution with hcl until it turns colourless

6. record the final burette reading of acid used

7. repeat until you have two concordant results

phenolphthalein colour change

pink alkali

colourless acid

methyl orange colour change

red in acid

yellow in alkali

what are concordant results

results within 0.1 of oneanother

why should the burette be rinsed with sodium hydroxide before filling rather than with water

so that the concentration of sodium hydroxide is not diluted by water

why do we do a rough tirtration

to get an idea of where the end point is