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Quantitative analysis Part B Stoichiometry

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Calculating % Composition in Compounds: Work out the Molar Mass (M) of these compounds: M (H2O)=

M (CO 2)=

This is the total mass of the compound. In a % composition calculation you are giving what % of the total mass is made up of the elements which make it up. Eg: What % of Oxygen is in H2O and CO2 % =

Mass of element

x100

Total mass of compound %O in H2O = 18

16 x 100 = 88.9% 44

%O in CO2 = 32 x 100 = 72.7%

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% CALCULATIONS 1(a) Calculate the molar mass of iron oxide, Fe2O3. Include units in your answer. (b)Calculate the percentage mass of iron in iron oxide. 2(a) Calculate the molar mass of aluminium oxide, Al2O3. Include units in your answer. (b)Calculate the percentage mass of oxygen in aluminium oxide. 3. Allicin is the compound responsible for the characteristic smell of garlic. Its formula is C6H10S2O. What is the % of oxygen in this compound?

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Steps for Determining an Empirical Formula then Molecular Formula Purple Monkeys Drink Rum – memory jog 1. (Percent to grams) Start with percentages of each element given, assume that the total mass is 100 grams so that the mass of each element = the percent given. 2. (Moles) Convert the mass of each element to moles using the molar mass from the periodic table. (using n = m/M) 3. (Divide by smallest) Divide each mole value by the smallest number of moles calculated. 4. (Round) Round to the nearest whole number. This is the mole ratio of the elements and is represented by subscripts in the empirical formula. e.g.  If one element ratio is 1.5, then multiply each element in the problem by 2 (to get 3) e.g.  If  one element ratio is 1.25, then multiply each element in the problem by 4 (to get 5). 

This is the Empirical formula’s ratio

eg The Question:

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a) Calculate the empirical formula of a compound which has a percentage composition of 52.2% carbon, 13.0% hydrogen, and 34.8% oxygen.

Following the Steps:

1.(Percent to grams) C= 52.2g    H= 13g   O= 34.8g 2. (moles) n (C) = 52.2 / 12 = 4.35

 3. (Divide by smallest) C= 4.35 /2.175 = 2 H= 13.0 / 2.175 = 5.98 O= 2.175 / 2.175 = 1 (This is the ratio of moles)

n (H) = 13.0 / 1 = 13.0

4. (Round) The 5.98 is so close to 6 we  can round it up to 6 which gives us the 

n (O) = 34.8 / 16 = 2.175

  Empirical formula of C2H6O

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Molecular formula calculation: 1. (M divided by E) Divide the actual molar mass of the compound (given in the  question) by the Molar Mass of the Empirical formula 2. (x E) The number you get is what you multiply the Empirical formula by. (From the previous example)

If the molar mass of this compound is 92 g/mol, use your answer to part (a) above to determine its molecular formula. Molecular formula calculation: (M divided by E) Empirical mass = 12 x 2 + 1 x 6 + 16 = 46 g/mol 92 (provided in Q) / 46 = 2 (x E)The number you get is what you multiply the Empirical formula by. so molecular mass is Empirical formula (C2H6O) x2 = C4H12O2 Try this link for more info http://sciencescribe.co.nz/files/12CHE/QUANT_empirical.pdf

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EMPIRICAL AND MOLECULAR FORMULA CALCS 1(a) Glucose is a sugar which has a mass composition of 40.0% carbon, 6.70% hydrogen, 53.3% oxygen. Calculate the empirical formula of glucose. (b) If the molar mass of glucose is 180 g/mol, use your answer to part (a) above to determine the molecular formula of glucose.

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EMPIRICAL AND MOLECULAR FORMULA CALCS contd 2(a) Caffeine is a stimulant which has a mass composition of 49.5% carbon, 5.20%, hydrogen, 28.9% nitrogen and 16.5% oxygen. Calculate the empirical formula of caffeine. (b)If the molar mass of caffeine is 194 g/mol, use your answer to part (a) above to determine the molecular formula of caffeine. -More practice: http:// sciencescribe.co.nz/files/12CHE/QUANT_empirical.pdf

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EMPIRICAL AND MOLECULAR FORMULA CALCS contd The cystine molecule forms an important part of protein structures. This cystine molecule is 40.91% carbon, 6.82% hydrogen, 36.36% oxygen and 15.91% nitrogen. (i) Calculate the empirical formula of cystine. (ii) If the molar mass of cystine is 176 g/mol calculate the molecular formula.

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Mole Ratios Calculations These are simply determined by looking at a balanced equation. The number before any of the reactants or products is the number of MOLES.   Eg:

2NaOH

+

H2SO4 

Na2SO4 +

2H2O

  In this equation the Mole Ratio of NaOH to H2SO4 is 2:1 (ie: you need twice the number of moles of NaOH to H 2SO4)   For the calculations: Get the ratio from the equation Unknown moles/known moles (U/K) x the given number of moles (n)

Mole Ratios Calculations

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2HCl + Zn(OH)2 → ZnCl2 + 2H2O n(HCl) = 0.5 mol;

2

n(ZnCl2) = ?

4Al + 3I2 a b

3

→

2AlI3

n(Al) = 0.6 mol

n(I2) = ?

n(AlI3) = 2.4 mol H2SO4 + 2KOH

n(I2) = ? →

K2SO4 + 2H2O

a

n(KOH) = 2.5 mol

n(H2SO4) = ?

b

n(K2SO4) = 10 mol

n(KOH) = ?

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RATIO CALCULATIONS (starting and finishing with a mass) Example Question: What mass of H2O is produced in the complete combustion of 25.4 g of methane according to the following equation?

CH4

+

2O2 → CO2

+

2 H2O

Known CH4 Mol ratio

1. Identify Known and Unknown

n

CH4 is the known (as we know there are 25.4g) and H 2O is the unknown as that is the mass we are trying to find out. THESE ARE THE COLUMN HEADINGS

m

2. Complete the table with the molar masses, mole ratio and known mass 3. Convert mass to moles for the KNOWN (using

n=m/M) n(CH4) = 25.4 g/ 16 = 1.5875 moles (known)

4. Use the MOLE RATIO to find the moles of UNKNOWN 1.5875 x 2 = 3.175 moles (of unknown)

5. Convert mass to moles for the UNKNOWN (using

n=m/M) m(H2O) = 3.175 moles x 18 gmol-1 = 57.15g = 57.2 g to 3sf.

Unknown H2 O

M SHOW WORKING OUTSIDE THE BOX! GIVE FINAL ANSWER TO 3SF! More infohttp://sciencescribe.co.nz/files/12CHE/Q UANT_massStoic.pdf More Q’s http://sciencescribe.co.nz/files/12CHE/ QUANT_massStoichQuestions.pdf

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RATIO CALCULATIONS:

1 What mass of H2O is produced in the complete combustion of 36.2 g of methanol according to the following equation? 2 CH3OH

+



3 O2

2

CO2

+

4 H 2O

2 What mass of carbon dioxide would be produced from the reduction of 6.20 kg of iron(III) oxide according to the following equation? Fe2O3

+

3 CO



2 Fe

+

3 CO2

3 What mass of CO2 is produced in the complete combustion of 34.5 g of ethanol according to the following equation? C2H5OH

+



3 O2

2 CO2

+

3 H 2O

4 What mass of iron can be produced from the reduction of 1.50 kg of iron (III) oxide according to the following equation? Fe2O3

+

3 CO



2 Fe

+

3 CO2

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Practice Assessment 1 1. A plant contains 1.5% of an acid which has 26.67% carbon,  2.22% hydrogen , 71.11% Oxygen and has a molar mass of  90.0 g mol1. Use the following molar masses to calculate both  the empirical formula and the molecular formula of the acid: M(C) = 12.0 g mol1 M(H) = 1.00 g mol1 M(O) = 16.0 g mol1  M(Na) = 23.0 g mol1 empirical formula calculation:

molecular formula calculation:

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Practice Assessment 1  2. The plant also contains small amounts of another acid –, formula C3H4O3. Calculate  the percentage composition (of the elements) in this acid.       3. This acid C3H4O3 reacts with sodium hydroxide. The equation for the reaction is: C3H4O3 + 2NaOH      C3H4O3Na2 + 2H2O Calculate the maximum mass of the sodium salt- C3H4O3Na2, which could be made from  26.0g of sodium hydroxide.

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Practice Assessment 2 1. A plant contains 1.5% of an acid which has 40.92% carbon, 4.58% hydrogen ,  54.5% Oxygen and has a molar mass of 176 g mol1. Use the following molar  masses to calculate both the empirical formula and the molecular formula of  the acid: M(C) = 12.0 g mol1 M(H) = 1.00 g mol1 M(O) = 16.0 g mol1 M(Na) = 23.0 g mol1 empirical formula calculation:  

molecular formula calculation:

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Practice Assessment 2 1. The plant also contains small amounts of another acid –, formula C2H4.  Calculate the percentage composition (of the elements) in this acid.

2. This acid C3H4O3 reacts with sodium hydroxide. The equation for the reaction is: C2H5O2 + 2NaOH      C2H5O2Na2 + 2H2O Calculate the maximum mass of the sodium salt- C2H5O2Na2, which could be made  from 33.0g of sodium hydroxide.

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Last minute tips before doing the titration assessment (Part B) • Always indicate what you are working out eg: n(C) = • Show all the units in your calculation after each number (you don’t need units for ratios or % ) • Consistently use 3sf in all of your FINAL answers for excellence • Check all your calculator work once again to make sure your data entry was correct

ANSWERS

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Slide 4 1(a)

M(Fe2O3) = 56 x 2 + 16 x 3 =

160 g/mol (b) 112 / 160 = 70% 2(a)

3

=47.1%

M(C6H10S2O) 162 g mol1 m (O) = 16 % O = 16 / 162 x 100

Slide 8

a) 40 / 12 = 3.3 1

b) m(CH2O) =

6.7 / 1 = 6.7 2

27 x 2 + 16 x 3 = 102gmol-1

(b) 48 / 102 = 0.529

Slide 8

30gmol-1 180 / 30 = 6 C6H12O6

53.3 / 16 = 3.3 1 Slide 9 a) empirical formula C4H5N2OCH2O Slide 9 b) Empirical mass = (4 x 12) + (5 x 1) + (2 x 14) + 16 = 97

/ 97 = 2 Slide 10: %n(C) 40.91 / 12 = 3.409 194 3 O ==9.87% n(H) = 6.82 / 1 = 6.82

molecular formula C8H10N4O2

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n(N) = 15.91 / 14 = 1.136

1

n(O) = 36.36 / 16 = 2.2725 2 C3H6NO2

so

Slide 12: Mole Ratio’s

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Q1 0.25 mol

Slide 14: Full ratio calcs Q1. n(CH3OH) = m / M = 36.2 / 32 = 1.13125 mol

Q2 a) 0.45 mol b) 3.6 mol Q3 a) 1.25 mol b) 20 mol

moles H2O = 2 x moles CH3OH = 2 x 1.13125 = 2.26 mol(3sf) (NB: Do your rounding to 3sf right at the end for the answer)  m(H2O) = n x M = 2.2625 x 18 = 40.7g

Q2. n(Fe2O3) = m / M = 6200 / 159.8 = 38.7985 moles moles CO2 = 3 x moles Fe2O3 = 3 x 38.7985 = 116 mol (3sf) m(CO2) = n x M = 116.39549 x 44 = 5121.4g = 5.12kg Q3. (i) n(C2H5OH) = 34.5 g / 46.0 g mol1 = 0.750 mol (ii)

n(CO2) = 2 x n(C2H5OH) = 1.50 mol

(iii)

m(CO2) = 1.50 mol x 44.0 g mol1 = 66.0 g

Q4. (i) n(Fe2O3) =1500 g / 160 g mol1 = 9.38 mol (ii)

n(Fe) = 2 x n(Fe2O3) = 18.8 mol

(iii)

m(Fe) = 18.8 mol x 55.9 g mol1 = 1050 g (1.05 kg)

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Practise Assessment 1 Slide 15-16 Q1. Empirical =CHO2 Molecular =C2H2O4   Q2. C = 40.9%

Practise Assessment 2 Slide 17-18 Q1. Empirical =C3H4O3

H= 4.6% O=54.5%

Q3. M (NaOH)=40 gmol-1 M(C3H4O3Na2) =134 gmol-1 n(NaOH)=26.0g/40=0.650mol   n(C3H4O3Na2 )U/K = ½ x 0.650=0.325mol m(C3H4O3Na2) = 0.325mol x 134 gmol-1=43.6g

Molecular =C6H8O6   Q2. C = 92.3%

H= 7.7%

Q3. M (NaOH)=40

gmol-1

M(C2H5O2Na2) =107 gmol-1

n(NaOH)=33.0g/40=0.825mol   n(C3H4O3Na2 )U/K = ½ x 0.825=0.413mol