**Topic 1 (1.1, 1.2 and 1.3)**

**Paper 1 style questions **are multiple choice. You are **not permitted to use a calculator or the data book** for these questions, but you should use a periodic table.

A **periodic table pop-up** is available on the left hand menu.

*Complete this question without a calculator.*

0.20 mol of hydrochloric acid is mixed with 0.20 mol of calcium carbonate:

2HCl + CaCO_{3} → CaCl_{2} + H_{2}O + CO_{2}

Which is correct?

Quantities of both reagents are given, which suggests that one reagent will be in excess and the other will be limiting. This is best considered in terms of the mole ratio:

2HCl | + | CaCO_{3} | → | CaCl_{2} | + | H_{2}O | + | CO_{2} |

2 | : | 1 | : | 1 | : | 1 | : | 1 |

2 moles of HCl will react with 1 mole of CaCO_{3}.

Therefore 0.20 moles of HCl will react with 0.10 moles of CaCO_{3}. 0.20 moles of CaCO_{3} has been used, so the calcium carbonate is in excess, and the hydrochloric acid is therefore the limiting reagent.

2 moles of HCl will yield 1 mole of CO_{2}.

Therefore 0.20 moles of HCl will yield 0.10 moles of CO_{2}.

Thus **Limiting reagent is HCl and maximum yield of CO _{2} is 0.10 mol **is the correct answer.

What is the sum of all coefficients when the following equation is balanced using the smallest possible whole numbers?

___GeO_{2} _{ }+ ___HCl → ___GeCl_{4} + ___H_{2}O

GeO_{2} + 4HCl → GeCl_{4} + 2H_{2}O_{ }is the balanced equation, so the total sum is 8 - don't forget that GeO_{2} and GeCl_{4} both have a '1' in front of them, although by convention we don't actually write that into the equation.

*Complete this question without a calculator.*

Which contains the greatest number of moles of oxygen atoms?

This question requires us first to work out the number of oxygen atoms in each formula, and to then multiply that by the number of moles, which will give us the number of moles of oxygen atoms:

0.05 mol Ca(NO_{3})_{2 }: calcium nitrate has 6 oxygen atoms: 6 × 0.05 = 0.30

0.10 mol C_{6}H_{4}(NO_{2})_{2 }: dinitrobenzene has 4 oxygen atoms: 4 × 0.10 = 0.40

0.20 mol CO : carbon monoxide has 1 oxygen atom: 1 × 0.20 = 0.20

0.10 mol O_{2} : oxygen has 2 oxygen atom: 2 × 0.10 = 0.20

Thus **0.10 mol C _{6}H_{4}(NO_{2})_{2} **is the correct answer as this contains the largest number of moles of oxygen atoms (0.40 mol).

Which of these is an example of a heterogeneous mixture?

A heterogeneous mixture is a mixture of more than one phase (or state) of matter. Hence **Solid calcium carbonate powder suspended in water **is the correct answer as this contains both a solid (calcium carbonate powder) and a liquid (water). The other answers are homogeneous (one phase of matter) mixtures (a solution like aqueous sodium chloride is uniform throughout and considered to be homogeneous), or in the case of nitrogen dioxide gas - a pure compound.

*Complete this question without a calculator.*

What is the empirical formula of a hydrocarbon with 90% carbon and 10% hydrogen by mass?

We need to find the simplest ratio of the number of (moles of) atoms. We cannot use a calculator so we will need to approximate the relative atomic masses that we take from the periodic table (e.g. carbon 12.01 = 12).

Start by assuming 100g: Thus 90g of C atoms, 10g of H atoms:

Carbon | Hydrogen | ||

moles = mass/molar mass | 90/12 = 7.5 | 10/1 = 10 | |

multiply up to whole numbers (×2) | 15 | 20 | |

find the largest common multiple to get the simplest ratio (5 in this case) | 15 ÷ 5 = 3 | 20 ÷ 5 = 4 |

Therefore the empirical formula and the correct answer is **C _{3}H_{4}**

Which relationship would give the following graph when plotted for a fixed mass of an ideal gas with all other variables constant?

The ideal gas equation is PV=nRT. The letters represent Pressure, Volume, number of moles, gas constant (R) and Temperature in that order.

P against T and V against T will both give a directly proportional linear relationship as the two variables are on opposite sides of the equals sign in the equation; so any change in one variable must be proportionally mirrored in the other variable if all the other values are constant.

P against \({1\over V}\) will also give a straight line. If all other values in PV=nRT are constant then PV=constant. If we give that constant a value of 1 (PV=1) then P =\({1 \over V}\) so volume is inversely proportional to pressure and P against \({1\over V}\) will give a straight line when plotted.

P against V will give the curve (a hyperbola) as shown in the diagram since (in the same way as above) PV=constant, as pressure is increased volume decreases and vica versa. This inversely proportional relationship will (mathematically) always give a curve when plotted.

(If the maths is too much - learn the laws (Avogadro's, Boyle's, Charles' and Dalton's) on the revision slide.)

Thus** P against V** is the correct answer.

*Complete this question without a calculator.*

How many moles of Iron will be produced if this reaction produces 500 mol of carbon dioxide?

Fe_{2}O_{3} + 3CO → 2Fe + 3CO_{2}

The mole ratio must be used to answer this question:

Fe_{2}O_{3} | + | 3CO | → | 2Fe | + | 3CO_{2} | ||

1 | : | 3 | : | 2 | : | 3 |

The reaction gives **2 **moles of iron (Fe) for every **3** moles of carbon dioxide (CO_{2}) produced.

Therefore for every 1 mole of CO_{2} produced, \({2 \over 3}\) of a mole of Fe is produced.

If 500 moles of CO_{2} are produced, \({2 \over 3}\) × 500 = 333 moles of Fe will be produced.

Thus **333 mol **is the correct answer.

What is the sum of the smallest integer coefficients when ethene undergoes complete combustion?

___C_{2}H_{4 }+ ___O_{2 }→ ___CO_{2} + ___H_{2}O

C_{2}H_{4 }+ 3O_{2 }→ 2CO_{2} + 2H_{2}O is the balanced equation, so the sum of the integer coefficients is **8** - don't forget that C_{2}H_{4} does have a '1' in front of it, although by convention we don't actually write that into the equation.

(When balancing combustion reactions of hydrocarbons, first balance out carbon and hydrogen atoms by adding coefficients (numbers) in front of CO_{2 }and H_{2}O, after that balance the oxygen atoms and the hydrocarbon as required.)

*Complete this question without a calculator.*

What is the volume of an ideal gas when the pressure on 100cm^{3} of gas is changed from 200kPa to 100kPa at constant temperature?

The ideal gas equation is PV=nRT. The letters represent Pressure, Volume, number of moles, gas constant (R) and Temperature in that order.

If all other values in PV=nRT are constant then PV=constant. If we give that constant a value of 1 (PV=1) then P =\({1 \over V}\) so volume is inversely proportional to pressure (and P against \({1\over V}\) will give a straight line when plotted).

Therefore, if we halve the pressure, from 200 to 100 kPa, we must double the volume: Thus** 200 cm ^{3} **is the correct answer.

*Complete this question without a calculator.*

What is the concentration, in mol dm^{−3}, of 10.0g of NaOH (M_{r} = 40.0) in 500cm^{3} of solution ?

**Concentration **(mol dm^{−3}) **= moles **(mol)** / volume of solution **(dm^{3})

**Moles **(mol) = **Mass **(g) / **molar mass **(g mol^{−1})

So moles of sodium hydroxide (NaOH) = 10.0/40.0 = 0.250

Concentration of the solution = 0.250 / 0.500 (The volume must be converted to dm^{3}: 1 dm^{3} = 1000cm^{3})

Concentration = 0.250 / 0.500 = **0.500 mol dm ^{−3}**

*Complete this question without a calculator.*

What volume of carbon dioxide can be obtained by reacting 500 cm^{3} of methane (CH_{4}) with 500 cm^{3 }of oxygen (O_{2})?

CH_{4(g)} + 2O_{2(g)} → CO_{2(g)} + 2H_{2}O_{(l)}

Ideal gases have a constant molar volume (at any given temperature and pressure).

Quantities of both reagents are given, which suggests that one reagent will be in excess and the other will be limiting. This is best considered in terms of the mole ratio:

CH_{4} | + | 2O_{2} | → | CO_{2} | + | 2H_{2}O | ||

1 | : | 2 | : | 1 | : | 2 |

1 mole of CH_{4} will react with 2 moles of O_{2}.

So methane will react with twice its volume of oxygen (and vice versa, oxygen will react with half its volume of methane).

Methane gas is therefore in excess and oxygen is the limiting reagent.

The amount of methane that will react is therefore 500 cm^{3} / 2 = 250cm^{3} (1:2 ratio)

And 250 cm^{3} of methane reacting with give 250 cm^{3} of carbon dioxide (CO_{2}) (1:1 ratio)

Thus the answer is **250cm ^{3}**

*Complete this question without a calculator.*

How many moles of CaSO_{4} are required to produce 128 g of SO_{2}? (Ar: S = 32, O = 16)

3CaSO_{4} + CaS → 4CaO + 4SO_{2}

The mole ratio must be used to answer this question:

3CaSO_{4} | + | CaS | → | 4CaO | + | 4SO_{2} | ||

3 | : | 1 | : | 4 | : | 4 |

The reaction produces **4 **moles of SO_{2} for every **3** moles of CaSO_{4} reacted.

Therefore for every 1 mole of SO_{2} produced, \({3 \over 4}\) of a mole of CaSO_{4} must be reacted.

Using **Moles **(mol) = **mass** (g) /** molar mass **(g mol^{−1}) (molar mass of SO_{2} = 32 + 16×2 = 64) 128g of SO_{2} is 128 / 64 = 2.0 moles

If 2.0 moles of SO_{2} is produced, \({3 \over 4}\) × 2.0 = 1.5 moles of CaSO_{4} will be required.

Thus **1.5 mol **is the correct answer.

*Complete this question without a calculator.*

24 g of bromine react with 18.1 g of metal, M, to form MBr_{3}. What is the relative atomic mass of metal M? (A_{r}: Br = 80)

The formula of the metal bromide is MBr_{3}. This tells us that one mole of metal atoms will combine with three moles of bromine atoms (1:3).

Using **Moles **(mol) = **mass **(g) / **molar mass **(g mol^{−1}), 24 g of bromine is 24 / 80 = 0.3 moles (the common multiple for mental arithmetic here is 8)

Using the ratio 3:1, 0.3 moles of bromine combines with 0.1 moles of metal, M.

So if 0.1 moles of metal M has a mass of 18.1 g (given in the question) then 1 mole of metal M will have a mass of 18.1 × 10 = 181 g

Therefore the relative atomic mass of metal M is **181**.

*Complete this question without a calculator.*

What is the molecular formula of a hydrocarbon containing 85.6% carbon by mass with an integer molar mass of 168g mol^{−1}?

We need to find the simplest ratio of the number of (moles of) atoms. We cannot use a calculator so we will need to approximate the relative atomic masses that we take from the periodic table (e.g. carbon 12.01 = 12).

Start by assuming 100g: Thus 85.6g of C atoms, 14.4g of H atoms:

Carbon | Hydrogen | ||

moles = mass/molar mass | 85.6/12 = A little over 7 (7×12=84) | 14.4/1 = 14.4 | |

Approximate mole ratio | 7... | 14... | |

Simplest mole ratio (empirical formula) | 1 | 2 |

This suggests that the empirical formula is probably CH_{2}. The empirical formula mass is therefore approximately 14 units (12 + 1+ 1).

168g mol^{−1} is the molar mass (rounded to a whole number) and 168 / 14 = 12

So there are 12 empirical formula units (of CH_{2}) in the molecular formula.

Therefore the molecular formula is **C _{12}H_{24}**.

(Once the approximate 1:2 ratio of C:H is known, the last part can also be done by trial and error: 12×12 = 144, 144+24 = 168)

*Complete this question without a calculator.*

What is the percentage yield (%) when 14 g of ethene produces 12 g of ethanol?

C_{2}H_{4} + H_{2}O → C_{2}H_{5}OH

Using Mr(ethene) = 28 and Mr(ethanol) = 46.

**Percentage yield** = (experimental yield / theoretical yield) × 100%

**Moles** = **mass **/ **molar mass**

Moles of ethene = 14 / 28

Moles of ethanol = 12 / 46

The reaction has a 1:1 molar ratio as seen in the equation. So the theoretical yield of ethanol in moles will be the same as the moles of ethene.

Therefore the percentage yield will be: (moles of ethanol / moles of ethene) × 100%

= (12 / 46) / (14 / 28) × 100%

(The easiest way to deal with this is probably to remember that dividing by a fraction is the same as multiplying by the inverted (upside down) fraction. So...)

= \({12\over 46}\) × \({28\over 14}\) × 100%

= \({12\times28\times100 \over 14\times46}\)** **

*Complete this question without a calculator.*

What is the value of x when 36.0g of MnCl_{2}.xH_{2}O is heated to dryness leaving 31.5g of anhydrous MnCl_{2}?

Using A_{r}: Mn = 55, Cl = 35.5, H = 1, O = 16

The value for the water of crystallisation, x, can be found by finding the integer ratio of **moles of MnCl _{2} : moles of H_{2}O**. (x is always given as a whole number.)

**Moles** (mol) = **mass **(g) / **molar mass **(g mol^{−1})

The moles of MnCl_{2} can be found using the mass of the anhydrous salt: 31.5 / 126 = \({1\over 4}\) = 0.25 mol

The mass of water can be found by subtracting the mass of anhydrous salt from hydrated salt: 36.0 − 31.5 = 4.5 g

And moles of water = 4.5 / 18 = \({1\over 4}\) = 0.25 mol

The ratio of **moles of MnCl _{2} : moles of H_{2}O** is therefore 0.25 : 0.25 and the integer ratio is therefore 1 : 1.

The value of x is therefore **1**.