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Moles and Equations

Moles and Equations

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Moles and Equations

ATOMIC SIZE AND ATOMIC MASSAtomic sizeThe atom is extremely small and chemists use a unit called the nanometre to measure such tiny objects.

1 nanometre, nm, is one-thousand-millionth of a metre (0.00000001 m.).

Measuring atomic massesThe mass of an atom is too small to be measured on a balance.

The carbon–12 isotope is chosen as the international standard for the measurement of atomic mass.

One atom of carbon–12 is defined as having a mass of exactly 12 u ‘u’ is the ‘unified mass unit’.1 u has a mass of one-twelfth the mass of one atom of carbon–12.1 u = 1.661 x 10–24 g

Problem: What is the mass of one atom of carbon-12?

A typical sample of carbon contains atoms of the carbon–12 isotope (98.89%), with small proportions of the heavier isotopes, carbon–13 and carbon–14.

Chemists measure atomic masses as relative masses This is done by comparing the masses of atoms with the mass of the carbon–12 isotope.

Relative isotopic massThe relative isotopic mass is the mass of an atom

of an isotope of an element compared to one-twelfth of the mass of the carbon–12 isotope.

The relative isotopic masses of the three carbon isotopes are shown below.

isotope relative mass12C exactly 1213C 13.0014C 14.00

The average mass of a carbon atom after taking into account the relative abundance of each isotope is 12.0111.


Moles and Equations

Relative atomic massChemists uses relative masses of atoms to compare the masses of the atoms of different elements.

The relative atomic mass, Ar, of an element is the weighted mean mass of an atom of the element compared to one-

twelfth of the mass of the carbon–12 isotope.

Weighted mean massThe weighted mean mass takes into account

the mass of each isotope

the percentage abundance of each isotope.

Bromine contains 53.0% of bromine-79 and 47.0% of bromine-81.

mass from 79Br = x ………

mean from 81Br = x ………

relative atomic mass (weighted mean mass) of bromine = ………….

ProblemsCalculate the relative atomic mass, Ar, of the following to 4 significant figures.

(a) Chlorine contains 75.53% 35Cl and 24.47% 37Cl.

(b) Iridium contains 38.5% 191Ir and 61.5% 193Ir.

(c) Silicon contains 92.18% 28Si, 4.71% 29Si and 3.12% 30Si.


Moles and Equations

MORE RELATIVE MASSESChemists use relative masses for all chemicals.

Relative molecular massThis is the relative mass of simple molecular substances, such as chlorine, Cl2, and water, H2O.

The relative molecular mass, Mr, is the weighted mean mass of a molecule of a compound compared to one-twelfth of the mass of the carbon–12 isotope.

The relative molecular mass can easily be calculated by adding together the relative atomic masses of the atoms within the molecule:

molecule Mr

Cl2: 35.5 x 2 = 71.0H2O: 1.0 x 2 + 16.0 = 18.0

Relative formula massThis is the relative mass of any formula unit. It is especially useful for ionic compounds, such as sodium chloride, NaCl, compounds that do not exist as simple molecules.

The relative formula mass is the weighted mean mass of the formula unit of a compound compared to one-

twelfth of the mass of the carbon–12 isotope.

Find a relative formula mass by simply add together the relative atomic masses of each element in the formula:

compound relative formula massNaCl: 23.0 + 35.5 = 58.5CaBr2: 40.1 + 79.9 x 2 = 199.9

Some relative questionsUse the relative atomic masses from the periodic table to calculate the relative formula mass of:

(a) HCl; (b) CO2; (c) H2S; (d) NH3; (e) Fe2O3;

(e) H2SO4: (f) HNO3; (g) NH4Br; (h) C6H12O6; (i) Pb(NO3)2.


Moles and Equations

THE MOLEChemists are able to count atoms using the mole concept.

The mole is the S.I. unit of amount of substance and the abbreviation for the mole is mol.

1 mole is defined as the amount of substance which contains as many particles as there are carbon atoms

in exactly12 g of the carbon–12 isotope.

The number of atoms in 1 mole of carbon–12 is a number called the Avogadro constant, NA.

The Avogadro constant is 602,000,000,000,000,000,000,000 per mole

or 6.02 x 1023 mol1.

To the nearest whole number,

12 g of carbon contains 1 mole of atoms (6.02 x 1023 atoms).

1 g of hydrogen contains 1 mole of atoms (6.02 x 1023 atoms).

207 g of lead contains 1 mole of atoms (6.02 x 1023 atoms).

Thus, for carbon,

12 g of carbon contains 1 mole of atoms (6.02 x 1023 atoms).

6 g of carbon contains 0.5 mol. of atoms (3.02 x 1023 atoms).

24 g of carbon contains 2 mol. of atoms (12.04 x 1023 atoms).


My name is: ‘Count Lorenzo

Romano Amedeo Carlo Avogadro di Quaregna e

Cerreto’

Moles and Equations

Just how big is 1 mole?The table below shows how the numerical value for Avogadro's Number has been modified through the years in the quest for more digits and accuracy for the number.

It is a miniature version of the quest for digits for PI: .

Year Value

1931 6.061 x 1023

1958 6.02 x 1023

1981 6.022045(31) x 1023

1993 6.0221367(36) x 1023

1 mole (6.02 x 1023) standard soft drink cans would cover the surface of the earth to a depth of over 200 miles.

If you had 1 mole (6.02 x 1023) of popcorn kernels, and spread them across the United States of America, the country would be covered in popcorn to a depth of over 9 miles.

If we were able to count atoms at the rate of 10 million per second, it would take about 2 billion years to count the atoms in one mole.


1 mole of marbles would cover the whole surface of the U.K. to a depth of 1,500 km

Moles and Equations

MOLES OF ATOMSRefer to the periodic table to complete the table below.

Element Symbol Relative atomic

mass, Ar

Number of

moles

Mass of element /g

Number of atoms

Hydrogen 1 2 2 g 12.04 x 1023

Carbon 1

Sulfur 2

Oxygen 3

Calcium 0.5

Sodium 46 g

Potassium 6.02 x 1023

Magnesium 3.01 x 1023

Gallium 34.85 g

Strontium 21.9 g

Bromine 18.06 x 1023

Arsenic 1.204 x 1023

Molybdenum 2.408 x 1023

Lithium 34.5 g

Rubidium 8.55 g

Silicon 5.62 g

Scandium 9 g

Cobalt 36.12 x 1023

Chromium 0.2

Nitrogen 0.01

Uranium-238 8

Phosphorus 0.31 g


Moles and Equations

MOLAR MASS, MThe molar mass of a substance is the mass of

one mole of the substance.

The units of molar mass are g mol1. This is a very useful term because it can be applied universally to elements and compounds.

The molar mass, M, is equal to the relative mass in grammes of the substance

For carbon, M(C) = 12.0 g mol1.

For water, M(H2O) = 18.0 g mol1.

Using molar masses,

number of moles =

In 4 g of C, amount (in moles) of C = = 0.33 mol

In 11 g of CO2, amount (in moles) of CO2 = = 0.25 mol

In 5.85 g of NaCl, amount (in moles) of NaCl = = 0.100 mol


Remember:

n =

Moles and Equations

Moles of molecules/formulaeRefer to the periodic table to complete the table below.

Compound Formula molar mass /g mol1

Number of moles

Mass of compound /

g

Number of molecules

Water H2O 18 1 18 g 6.02 x 1023

Ammonia NH3 1

Carbon dioxide 0.5

Carbon disulfide

7.62 g

Phosphine PH3 68 g

Methane CH4 64 g

Methanol CH3OH 3.01 x 1023

Dichlorine oxide

Cl2O 3

Hydrogen fluoride

12.04 x 1023

Acetone C3H6O 29 g

Hydrogen bromide

18.06 x 1023

Nitrogen trichloride

0.1

Ethane C2H6 0.25

Ethanol 1.5

Benzene C6H6 468 g

Phosphorus pentachloride

PCl5 10

Octane C8H18 348 g

PCl3 0.2

Silane SiH4 6.02 x 1024

Glucose C6H12O6 6.02 x 1022


Moles and Equations

More Mole questions

1. Calculate the mass of each of the following

(a) 1 mole of hydrogen molecules, H2

(b) 1 mole of hydrogen atoms, H

(c) 1 mole of silicon dioxide, SiO2

(d) 0.5 mole of carbon dioxide, CO2

(e) 0.25 mole of hydrated sodium carbonate Na2CO3•10H2O.

2. How many moles are each of the following?

(a) 32 g of oxygen molecules, O2

(b) 32 g of oxygen atoms, O

(c) 31 g of phosphorus molecules, P4

(d) 32.1 g of sulfur molecules, S8

(e) 50.05 g of calcium carbonate, CaCO3.


Moles and Equations

TYPES OF CHEMICAL FORMULAEmpirical formula

The empirical formula of a compound is the simplest, whole-number ratio of elements in a compound.

An empirical formula is often calculated from experimental results using the mole concept.

Worked Example 1In an experiment, 10.025 g of calcium combined with 9.5 g of fluorine.[Ar: Ca, 40.1; F, 19.0.]

molar ratio of atoms Ca : F

= :

…….. : ……..

divide by smallest number…….. : ……..

Empirical formula: ……..

Worked Example 2Analysis of a compound produced the percentage composition by mass of potassium: 70.9%; sulfur: 29.1%. [Ar: K, 39.1; S, 32.1.]

100.0 g of the compound contains 70.9 g of potassium and 29.1 g of sulfur

molar ratio of atoms K : S

= :

…… : …….

divide by smallest number …… : …….

Empirical formula: ……


Moles and Equations

Formula Determination1. Find the formula of the following:

(a) calcium oxide (4.01 g of calcium reacts to form 5.61 g of calcium oxide).

(b) sodium oxide (2.3 g of sodium reacts to form 3.1 g of sodium oxide).

(c) iron oxide (11.16 g of iron reacts to form 15.96 g of iron oxide).

2. Use the following percentage compositions by mass to find the empirical formula of the compound.(a) A compound of iron and bromine [Fe, 18.9 %; Br, 81.1 %].

(b) A compound of sulfur and oxygen [S, 40.0 %; O, 60.0 %]

(c) A compound of iron, sulfur and oxygen [Fe, 36.8 %; S, 21.1 %; O, 42.1 %]


Moles and Equations

Molecular formulaThe molecular formula of a compound is the actual

number of atoms of each element in a molecule.

A molecular formula is determined by comparing the empirical formula with the relative molecular mass, Mr, of the compound.

Worked ExampleA compound has an empirical formula of CH2 and a relative molecular mass, Mr, of 56. What is its molecular formula?

empirical formula mass of CH2: 12 + 1 x 2 = 14

number of CH2 units in a molecule: = 4

molecular formula: C4H8

Questions1. An empirical formula of CH2 can apply to different molecules, depending on Mr.

Complete the table below to find the different molecular formulae from CH2.

empirical formula empirical formula mass Mr multiple molecular formula

CH2 14

28 x 2 C2H4

42

70

112

154

322


Moles and Equations2. Use the following information to find the empirical and molecular formula of

each compound.

(a) A compound of carbon and hydrogen formed when 2.4 g of carbon combines with 0.6 g of hydrogen, Mr 30.

(b) A compound of carbon and hydrogen and oxygen formed when 0.3 g of carbon combines with 0.1 g of hydrogen and 0.4 g of oxygen, Mr 32.

(c) A compound of carbon and hydrogen and oxygen with the composition by mass: C, 40.0 %; H, 6.7 %; O, 53.3 %; Mr 180.


Moles and Equations

WATER OF CRYSTALLISATIONThe familiar blue crystals of ‘copper sulfate’ owe their colour to the presence of water of crystallisation. When copper sulfate crystals are heated they turn a paler blue until all the water is removed. You are left with just a white powder.

Hydrated describes the crystalline compound which contains water molecules in its structure.

Anhydrous describes the compound when all the water of crystallisation has been removed.

When compounds are crystallised from water, they frequently contain water molecules within their structure giving a crystal appearance to the compound. This water is known as the water of crystallisation.

We can work out the amount, in moles, of water in a crystal structure using percentage or mass composition data. The method is similar to that used in empirical formula calculations.

Example 1:A sample of 7.392 g of MgSO4 crystals lost 3.780 g of water on heating.

Calculate n in the formula MgSO4•nH2O.

molar ratio MgSO4 : H2O

= :

…….. : ……..divide by smallest number …….. : ……..

Formula: ……..……..

Example 2A sample of 9.98 g of CuSO4•xH2O was heated until the blue crystals became colourless once all of the water had been removed. The residue was re-weighed and the mass was found to be 6.38 g.

Calculate x in the formula CuSO4•xH2O.

molar ratio CuSO4 : H2O

= :

…….. : ……..divide by smallest number …….. : ……..

Formula: ……..……..


Notice the dot in the middle of the formula.

Moles and Equations

Questions1. From the experimental results below, work out the formula of the hydrated salt.

Mass of BaCl2•xH2O = 7.329 gMass of BaCl2 = 6.249 g

2. 1.1895 g of CoCl2•xH2O was heated to drive off all of the water of crystallisation. The mass of the anhydrous cobalt chloride was 0.6495 g.

Calculate the value of x to determine the formula of the hydrated salt.

3. 12.9375 g of ZnSO4•xH2O was heated to drive off all of the water of crystallisation. The mass of the anhydrous zinc sulfate was 7.2675 g.

Calculate the value of x to determine the formula of the hydrated salt.


Moles and Equations

EQUATIONSChemical reactions involve the rearrangement of atoms and ions.

Chemical equations provide two types of information :

qualitative - which atoms or ions are rearranging

quantitative - how many atoms or ions are rearranging.

Qualitative informationAt its simplest qualitative level, a word equation can summarise a reaction:

e.g. hydrogen + oxygen water

The formula of each substance provides information about the chemicals that are reacting together:

H2 + O2 H2O

State symbols can be added to provide information about the physical states of each species under the conditions of the reaction :

gaseous state (g)liquid state (l)solid state (s)aqueous solution (aq)

H2(g) + O2(g) H2O(l)

Balancing an equation accounts for all of the species in the reaction.

In a correctly balanced equation, there will be the same number of each element on each side of the equation:


Moles and Equations

Balancing equationsBalance each of the following equations:

1. Fe(s) + S(s) FeS(s)

2. Na(s) + Cl2(g) NaCl(s)

3. N2(g) + O2(g) NO2(g)

4. H2(g) + Cl2(g) HCl(g)

5. Mg(s) + N2(g) Mg3N2(s)

6. K(s) + O2(g) K2O(s)

7. Al(s) + O2(g) Al2O3(s)

8. S(s) + O2(g) SO3(g)

9. P4(s) + O2(g) P2O5(s)

10. Fe(s) + Cl2(g) FeCl3(s)

11. Ca(s) + HCl(aq) CaCl2(aq) + H2(g)

12. KI(aq) + Cl2(g) KCl(aq) + I2(aq)

13. NaOH(aq) + H2SO4(aq) Na2SO4(aq) + H2O(l)

14. Al(s) + HCl(aq) AlCl3(aq) + H2(g)

15. CH4(g) + O2(g) CO2(g) + H2O(l)

16. N2(g) + H2(g) NH3(g)

17. MgCO3(s) + HCl(aq) MgCl2(aq) + H2O(l) + CO2(g)

18. NH3(aq) + H2SO4(aq) (NH4)2SO4(aq)

19. Fe(s) + H2SO4(aq) Fe2(SO4)3(aq) + H2(g)

20. Al2O3(s) + H2SO4(aq) Al2(SO4)3(aq) + H2O(l)


Balancing is done by trial and error

http://www.clipart.com/en/close-up?o=3770949&memlevel=A&a=a&q=balancing&k_mode=all&s=1&e=18&show=&c=&cid=&findincat=&g=&cc=&page=&k_exc=&pubid=%00%E5%A1%B9%EF%92%81%E1%B4%BB%E4%A1%BF%E2%B2%AF%E5%B6%82%E8%97%84%E6%8C%A7%00%00%EA%AE%A5

Moles and Equations

EQUATIONS AND THE MOLEThe balancing number is the amount of each substance in moles:

equation 2 H2(g) + O2(g) 2 H2O(l)

moles … mol ... mol … molreacting masses ………. g ………. g ………….. g

….… g ……..g ……..g

……. g of H2(g) reacts with …….. g O2(g) to form ……… g of H2O(l)

Now work out the masses of reactants and products from each of the balanced equations below.

1. 2 SO2(g) + O2(g) 2 SO3(g)

2. Mg (s) + 2 HCl(aq) MgCl2(aq) + H2(g)

3. 4 Al(s) + 3 O2(g) 2 Al2O3(s)


Moles and Equations

Working out reacting quantitiesWe can use this idea to work out quantities involved in any reaction. It is often better to work in moles right from the start.

ExampleNitrogen and oxygen react to form nitrogen monoxide, NO.

N2(g) + O2(g) 2 NO(g)

Calculate the masses of N2 and O2 required to form 3 g of NO.

STAGE 1

Work out the number of moles of NO that is formed.

n(NO) = = = ……… mol

n = number of moles; m = mass; M = Molar mass

STAGE 2

Write down a BALANCED EQUATION to find the moles used in this reaction.

We do this by scaling the balancing numbers to match 0.1 mol NO.

STAGE 3

Find the masses of N2(g) and O2(g) from their moles.

n(NO) = m = n x M

mass of N2 = …….. x ……….. = …………….g

mass of O2 = …….. x ……….. = …………….g

Some quantitative questions1. For the reaction: H2(g) + F2(g) 2 HF(g)


equation N2(g) + O2(g) 2 NO(g)

moles ............. mol + ............. mol ............. mol

moles used ............. mol ............. mol ............. mol

Remember:

n =

Moles and Equations Calculate the masses of hydrogen and fluorine required to form 5 g of hydrogen

fluoride, HF.

2. For the reaction: N2(g) + 3 H2(g) 2 NH3(g)

Calculate the mass of ammonia, NH3 that could be formed by reacting 30 g of hydrogen with an excess of nitrogen.

3. For the reaction: 4 Na(s) + O2(g) 2 Na2O(s)

Calculate the masses of sodium and oxygen required to form 5.25 g of sodium oxide, Na2O. Give your answers to 3 significant figures.


Moles and Equations

GASES AND THE MOLEIn 1811, Avogadro put forward a hypothesis:

In English:Equal volumes of gases contain the same number of molecules under the same conditions of temperature

and pressure.or

1 mole of molecules of any gas occupies 24.0 dm3 (24,000 cm3) at room temperature and pressure (RTP).

Working out moles from gas volumesWe can work out the number of moles of gas molecules present by simply measuring the volume:

For a gas at room temperature and pressure,

volume in dm3: Number of moles =

volume in cm3: Number of moles =

Worked ExampleHow many moles of gas molecules are in 480 cm3 of a gas at RTP?

Number of moles =

Number of moles = ……..…. = ………. mol


Moles and Equations

Questions1. How many moles of gas molecules are in the following gas volumes at r.t.p.

(a) 96 dm3 ………… (b) 3600 dm3 ………… (c) 2.0 dm3 …………

(d) 24 cm3 ………… (e) 2 cm3 ………… (f) 1.6 cm3 …………

2. What is the volume of the following moles of gas at r.t.p.

(a) 10 mol ………… (b) 75 mol ………… (c) 0.4 mol …………

(d) 0.006 mol ………… (e) 0.0025 mol ………… (f) 1.82 x 105 mol …………

3. What is the volume, at r.t.p., of

(a) 44 g CO2(g);

(b) 7 g N2(g);

(c) 5.1 g NH3(g)?

4. What is the mass, at r.t.p., of

(a) 1.2 dm3 O2;

(b) 720 cm3 CO(g);

(c) 48 cm3 CH4(g)?


n =

n =

Moles and Equations

Calculations involving gas volumesAs with masses, we can use the mole concept with gas volumes.

ExampleHydrogen peroxide, H2O2, decomposes into water and oxygen.

2 H2O2(l) 2 H2O(l) + O2(g)What volume of oxygen, at RTP, is formed by the decomposition of 17 g of hydrogen peroxide, H2O2(l)?

STAGE 1

n(H2O2) = = = ……… mol

n = number of moles; m = mass; M = Molar mass

STAGE 2

Write down a BALANCED EQUATION to find the moles used in this reaction.

We do this by scaling the balancing numbers to match 0.5 mol H2O2.

equation 2 H2O2(l) 2 H2O(l) + O2(g)

moles 2 mol H2O2(l) 2 mol H2O(l) + 1 mol O2(g)

moles used 0.5 mol H2O2(l) 0.5 mol H2O(l) + 0.25 mol O2(g)

STAGE 3

Find the volume of O2 from its moles.

n = V = n x 24 000

volume of O2 = …….. x ……….. = …………….cm3


n = =

Moles and Equations

More quantitative questions1. For the reaction: 2NaNO3(s) 2NaNO2(s) + O2(g)

What volume of O2, in cm3, is formed at RTP from 4.25 g of NaNO3?

2. For the reaction: 2CO(g) + O2(g) 2CO2(g)

What volumes, in dm3, of CO and O2 react together to form 13.2 g of CO2?

4. Mg was reacted with O2 to form 1.612 g MgO

(a) Write a balanced equation for the reaction

(b) What mass of Mg was reacted?

(c) What volume, in cm3, of O2, was reacted at RTP?


Moles and Equations2. Balance the equation: Al(s) + Cl2(g) AlCl3(s)

(a) What mass of AlCl3 forms by reacting 5.4 g of Al with an excess of Cl2?

(b) 9.0 g of Al reacted with 9 dm3 of Cl2 at RTP.Which reactant was in excess?

5. A compound of nitrogen and oxygen, A, was formed by reacting 20 cm3 of nitrogen, gas with 10 cm3 of oxygen.

(a) Determine the formula of compound A.

(b) Write an equation for this reaction.

(c) What volume of compound out formed at r.t.p.?

(d) What volumes of nitrogen and oxygen would have reacted to form 30 cm3

of a gaseous oxide of nitrogen with the formula

(i) NO2;

(ii) NO?


Moles and Equations

SOLUTIONS AND THE MOLEThe concentration of a solution is usually expressed in the moles of dissolved substance per cubic decimetre, (1000 cm3) of solution.

1 dm3 is equivalent to 1 litre, l.

1 cm3 is equivalent to the millilitre, ml.

A concentration of 1 mol dm–3 contains 1 mole of dissolved solute in 1 dm3 of solution.

A concentration of 2 mol dm–3 contains 2 moles of dissolved solute in 1 dm3 of solution.

Solutions of sodium chloride, NaCl(aq)1 mole of NaCl has a mass of 23.0 + 35.5 = 58.5 g.

A 1 mol dm–3 aqueous solution of NaCl contains 58.5 g of NaCl dissolved in sufficient water to make 1 dm3 of an aqueous solution.

A 2 mol dm–3 aqueous solution of NaCl contains 2 x 58.5 g = 117.0 g per cubic decimetre of aqueous solution.

Working out moles from volumes of solutionsProvided we know the molar concentration of a solution, we can work out the number of moles by measuring the volume:

For a solution of concentration c mol dm3,

volume in dm3: Number of moles = c x V (in dm3)

volume in cm3: Number of moles = c x


n = c x V (in dm3)

n = c x

Moles and Equations

Concentration questions1. Find the number of moles dissolved in:

(a) 1 dm3 of a 1 mol dm–3 solution,

(b) 2 dm3 of a 1 mol dm–3 solution,

(c) 4 dm3 of a 5 mol dm–3 solution,

(d) 250 cm3 of a 1.00 mol dm–3 solution,

(e) 10 cm3 of a 2.0 mol dm–3 solution,

(f) 20cm3 of a 0.10 mol dm–3 solution

(g) 26.2 cm3 of a 0.500 mol dm–3 solution,

(h) 29.6 cm3 of a 0.123 mol dm–3 solution,

(i) 13.2 cm3 of a 0.0260 mol dm–3 solution,

(j) 7.2 cm3 of a 0.0032 mol dm–3 solution.

2. Find the concentration in mol per cubic decimetre of solution for:

(a) 2 moles dissolved in 1 dm3 of solution,

(b) 4 moles dissolved in 2 dm3 of solution,

(c) 0.500 moles dissolved in 250 cm3 of solution,

(d) 2.00 moles dissolved in 200 cm3 of solution,

(e) 0.0100 moles dissolved in 100 cm3 of solution.

3. Find the concentration in grams per cubic decimetre of solution for:

(a) 2 moles of NaOH dissolved in 1 dm3 of solution,

(b) 4 moles of HCl dissolved in 2 dm3 of solution,

(c) 0.500 moles of H2SO4 dissolved in 250 cm3 of solution,

(d) 2.00 moles of HNO3 dissolved in 200 cm3 of solution,

(e) 0.0100 moles of Ca(OH)2 dissolved in 100 cm3 of solution.


Moles and Equations

CALCULATIONS FROM TITRATIONSThe manipulation of titration results follow a set pattern.

There are 3 stages involved and these MUST be learnt.

Worked exampleIn a titration 25.0 cm3 of 0.100 mol dm–3 aqueous sodium hydroxide were found to react with 10.4 cm3 of sulfuric acid. Find the concentration of the sulfuric acid.

STAGE 1

Work out the number of moles of WHAT YOU CAN.

Number of moles of NaOH = c x mol

= …………….. mol

= ……….. mol NaOH

STAGE 2

Write down a BALANCED EQUATION to find out the number of moles of the other reactant.

2 NaOH(aq) + H2SO4(aq) Na2SO4(aq) + 2 H2O(l)

…………mol ………. mol (balancing numbers)

……….. mol ……… mol

STAGE 3

Find the concentration of H2SO4 from its moles and volume.

n = c x c = n x


concentration of H2SO4 = …….. x = ……………. mol dm–3

Moles and Equations


Moles and EquationsProblems to concentrate the mind

1. In a titration 25.0 cm3 of 0.250 mol dm–3 aqueous sodium hydroxide were found to react exactly with 27.8 cm3 of hydrochloric acid. Find the concentration of the hydrochloric acid.

HCl(aq) + NaOH(aq) NaCl(aq) + H2O(l)

2. In a titration 25.0 cm3 of 0.200 mol dm–3 aqueous potassium hydroxide were found to react exactly with 15.4 cm3 of hydrochloric acid. Find the concentration of the hydrochloric acid.

HCl(aq) + KOH(aq) KCl(aq) + H2O(l)

3. In a titration 25.0 cm3 of 0.500 mol dm3 aqueous sodium hydroxide were found to react exactly with 23.2 cm3 of nitric acid. Find the concentration of the nitric acid.

HNO3(aq) + NaOH(aq) NaNO3(aq) + H2O(l)


Moles and Equations4. In a titration 25.0 cm3 of 1.00 mol dm3 aqueous sodium hydroxide were found to

react exactly with 26.6 cm3 of sulfuric acid. Find the concentration of the sulfuric acid.

H2SO4(aq) + 2 NaOH(aq) Na2SO4(aq) + 2 H2O(l)

5. A standard aqueous solution of sodium hydroxide was prepared containing 1.00 g in 250 cm3 of solution. In a titration 25.0 cm3 of this solution were found to react exactly with 20.5 cm3 of hydrochloric acid. Find the concentration of the standard solution and hence the concentration of the hydrochloric acid.

HCl(aq) + NaOH(aq) NaCl(aq) + H2O(l)

6. A standard aqueous solution of sodium carbonate was prepared containing 10.6 g in 100 cm3 of solution. In a titration 25.0 cm3 of this solution were found to react exactly with 11.3 cm3 of hydrochloric acid. Find the concentration of the standard solution and hence the concentration of the hydrochloric acid.

2 HCl(aq) + Na2CO3(aq) 2 NaCl(aq) + H2O(l) + CO2(g)


Moles and Equations


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