1.1 Matter, Chemical Change & the Mole Concept

Carvone is an organic compound containing carbon hydrogen and oxygen.

Complete combustion of 0.1526 g carvone produces 0.4470 g of carbon dioxide and 0.1281 g of water. Determine the empirical formula of carvone, showing your working.

0.146 g sample of carvone, when vaporised, had a volume of 0.0341 dm³ at 150 ^oC and 100.2 kPa. Calculate its molar mass showing your working.

Using your answer to part b), determine the number of molecules of carvone in 0.146 g.

Using Section 4 of the Data booklet, calculate the mass, in g, of protons in 38.46 g of ammonium sulfate.

The coca cola company states that a 12 oz can of coke contains 34 mg of caffeine, M = 194.19 g mol^-1. Calculate the number of moles of caffeine in a can of coke.

The composition by mass of caffeine is 49.48%  carbon, 5.20% hydrogen, 28.85% nitrogen and the rest is oxygen. Calculate the empirical formula of caffeine.

Calculate the molecular formula of caffeine using your answer from part b).

State the type of mixture coca cola is.
Does this change after the can of coke is opened? Justify your answer.

Concentrated nitric acid can be made to react with sulfur, and will produce sulfuric acid, nitrogen dioxide and water as the only products. Write a balanced equation for the reaction.

Deduce the redox changes taking place in the reaction in part a). 

Examine the state changes in the reaction in part a) and suggest the relative difference in the energy content of the products and reactants.

Identify the ions in part a) and calculate the sum of the relative formula masses of all the ions in reaction equation, using section 6 of the Data booklet.

The open-chain structure of the sugar galactose is shown in Figure 1 below.

 Figure 1

Determine the empirical formula of galactose.

Calculate the percentage composition by mass of galactose.

Formulate the balanced equation for the complete combustion of galactose. 

Lactose, C₁₂H₂₂O₁₁, is a disaccharide sugar made from galactose and glucose unit joined together. A number of people suffer from lactose intolerance in their diet and research shows that the vast majority of sufferers can tolerate up to 12 g with few symptoms.

How many molecules of lactose is enough to cause symptoms?

Common salt, or sodium chloride, can be used on icy roads in the winter to lower the freezing point of water. However, there is a limitation to using it, as the lowest freezing point that can be reached using sodium chloride is about -21 ^oC at 23% by mass of sodium chloride.

Other salts such as calcium chloride can also be used for this purpose. Figure 1 below shows a freezing point graph for calcium chloride/water.

Estimate the freezing point of water when the composition is 23% by mass of calcium chloride.

Figure 1

What % mass of calcium chloride would be needed to lower the freezing point of water in a 5-litre bucket to -30 ^oC and what would be the density of the resulting solution?

Antifreeze is a chemical largely consisting of ethylene glycol, C₂H₆O₂, which also lowers the freezing point of water and is used in car cooling systems. Figure 2 shows the freezing point graph for a water/antifreeze mixture.

Figure 2

If a car cooling system holds 5 litres of water, what is the minimum number of molecules of ethylene glycol needed to lower the freezing point to -50 ^oC?

The density of ethylene glycol is 1.11 gcm^-3.

Suggest one reason why is it better to use ethylene glycol in a car's cooling system rather than calcium chloride.

Camphor is a waxy, flammable, transparent solid with a strong aroma. It contains only carbon, hydrogen, and oxygen. It is found in the wood of the camphor laurel, a large evergreen tree found in East Asia.

Combustion analysis of camphor showed that a 2.450 g sample of camphor when burned in excess oxygen, produced 7.081 g of carbon dioxide and 2.320 g of water. Deduce the empirical formula of camphor.

Deduce the molecular formula of camphor (M = 152.23 gmol^-1).

Formulate a balanced equation, including state symbols, for the complete combustion of camphor.

Calculate the number of molecules in a teaspoon of camphor. The teaspoon holds 5.0 g of camphor (M = 152.23 gmol^-1).