Topic 1.3 - Formulae, equations and amounts of substance Lesson 3 - Chemical calculations e. use chemical equations to calculate reacting masses and vice versa using the concepts of amount of substance and molar mass f. use chemical equations to calculate volumes of gases and vice versa using the concepts of amount of substance and molar volume of gases, eg calculation of the mass or volume of CO 2 produced by combustion of a hydrocarbon (given a molar volume for the gas) g. use chemical equations and experimental results to deduce percentage yields and atom economies in laboratory and industrial processes and understand why they are important Connector: Write the formula for the following compounds and then calculate their molar mass. Lithium chloride, silver(I) nitrate, lead(II) iodide, aluminium fluoride, chromium(III) oxide magnesium nitrate Crowe 2011
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Topic 1.3 - Formulae, equations and amounts of substance
Topic 1.3 - Formulae, equations and amounts of substance. Lesson 3 - Chemical calculations. e. use chemical equations to calculate reacting masses and vice versa using the concepts of amount of substance and molar mass - PowerPoint PPT Presentation
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Topic 1.3 - Formulae, equations and amounts of substanceLesson 3 - Chemical calculations
e. use chemical equations to calculate reacting masses and vice versa using the concepts of amount of substance and molar massf. use chemical equations to calculate volumes of gases and vice versa using the concepts of amount of substance and molar volume of gases, eg calculation of the mass or volume of CO2 produced by combustion of a hydrocarbon (given a molar volume for the gas)g. use chemical equations and experimental results to deduce percentage yields and atom economies in laboratory and industrial processes and understand why they are important
Connector:
Write the formula for the following compounds and then calculate their molar mass.
Calculating masses from balanced equationsEquations can be used to tell us how much of a chemical is reacting or is produced.
ExampleWhat mass of magnesium oxide would be produced from 16g of magnesium in the reaction between magnesium and oxygen?i Write out the full balanced equation, including state symbols;
2Mg (s) + O2 (g) → 2MgO (s)
ii Read the equation in terms of moles: 2 moles of magnesium reacts to give 2 moles of magnesium oxide
Note: The numbers written in front of the substances in an equation represent moles, where 1 mole = formula weight in grams.
iii Convert the moles to masses using their relative atomic mass (Mr) values
2Mg (s) + O2 (g) 2MgO (s)
(2x 24g) 2x (24 + 16) = 48g = 80g
iv Use the answers to calculate the mass of MgO made from 16g of Mg From (iii) 48g of Mg would make 80g of MgO So 16g would make (80 x 16) / 48 = 26.7g
Example 2What mass of lead(II) sulphate would be produced by the action of excess dilute sulphuric acid on 10 g of lead nitrate dissolved in water?
1 mole gives 1 mole 331.2g gives 303.2g 10 g gives (303.2 x10) / 331.2 = 9.15 g
Example 3
Example 3
ethanol
Atom economy• The atom economy of a chemical reaction is a measure of the
amount of starting materials that become useful products.• Inefficient, wasteful processes have low atom economies.• Efficient processes have high atom economies, and are
important for sustainable development, as they use fewer natural resources and create less waste.
• The atom economy of a reaction can be calculated:
Note that, because the total mass of products equals the total mass of reactants, you can put that into the bottom of the fraction in the calculation like this:
Summary questions
1. Limestone, when it is heated, decomposes to form the commercially important material quicklime and the waste gas, carbon dioxide:
CaCO3 CaO + CO2
a) What are the formula masses of the above compounds?
b) What is the % of Ca in CaCO3 and CaO?
c) What is the atom economy of the above reaction?
2. Nitrogen and hydrogen are used to make ammonia:
N2 + 3H2 2NH3
a) What is the % of N in NH3?b) What is the atom economy of this reaction?
The molar volume of a gas is the volume occupied by one mole at room temperature and atmospheric pressure (r.t.p). It is equal to 24 dm3 or 24000 cm3 at r.t.p.
The molar volume of a gas is the volume occupied by one mole at room temperature and atmospheric pressure (r.t.p). It is equal to 24 dm3 or 24000 cm3 at r.t.p.