Formulas (Cambridge (CIE) A Level Chemistry): Flashcards
Exam code: 9701
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Define compound ion.
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Cards in this collection (36)
Define compound ion.
A compound ion is an ion that contains more than one type of element, such as hydroxide (OH-), sulfate (SO42-) or nitrate (NO3-).
What is the formula of magnesium chloride? Explain how you determine it.
The formula is MgCl2. Magnesium (Group 2) has a 2+ charge and chlorine (Group 17) has a 1- charge. Two Cl- ions are needed to balance one Mg2+ ion.
True or False?
Ionic compounds are electrically neutral because the total positive charge equals the total negative charge.
True.
In any ionic compound, the charges from positive and negative ions must balance exactly, giving an overall charge of zero.
Elements in Group 17 gain .......... electron(s) to form ions with a .......... charge, while elements in Group 16 gain .......... electrons to form ions with a 2- charge.
Elements in Group 17 gain 1 electron(s) to form ions with a 1- charge, while elements in Group 16 gain 2 electrons to form ions with a 2- charge.
What is the formula of iron(III) oxide? Explain how you determine it.
The formula is Fe2O3. Iron(III) has a 3+ charge and oxygen (Group 16) has a 2- charge. Two Fe3+ ions (6+) balance three O2- ions (6-).
Define Roman numerals in compound names.
Roman numerals in compound names are symbols used to indicate the charge (oxidation state) of the metal ion. They are used for transition metals whose charges can vary. For example, copper(II) has a 2+ charge.
True or False?
Ammonium (NH4+) is an example of a non-metal positive ion.
True.
Most positive ions are metals, but ammonium (NH4+) and hydrogen (H+) are non-metal positive ions.
The formula of aluminium nitrate is .......... because aluminium has a .......... charge and nitrate (NO3-) has a 1- charge.
The formula of aluminium nitrate is Al(NO3)3 because aluminium has a 3+ charge and nitrate (NO3-) has a 1- charge.
What charges do Group 1, Group 2 and Group 13 metals form when they become ions?
Group 1 metals form 1+ ions, Group 2 metals form 2+ ions and Group 13 metals form 3+ ions.
Define spectator ions.
Spectator ions are ions present in solution that do not take part in a chemical reaction. They appear on both sides of the full ionic equation and are cancelled out when writing the net ionic equation.
What are the four state symbols used in chemical equations and what do they represent?
(s) = solid, (l) = liquid, (g) = gas, (aq) = aqueous (dissolved in water).
True or False?
When balancing an equation, you can change the formulae of reactants or products to make the numbers work.
False.
You must never change the formulae. Only the coefficients (numbers placed in front of formulae) may be changed when balancing an equation.
The balanced equation for magnesium burning in oxygen is .......... Mg (s) + O2 (g) → .......... MgO (s).
The balanced equation for magnesium burning in oxygen is 2 Mg (s) + O2 (g) → 2 MgO (s).
Define ionic equation.
An ionic equation is a chemical equation showing only the ions and particles that actually take part in a reaction. Spectator ions are omitted and are cancelled from both sides of the full ionic equation.
Write the net ionic equation for the reaction of zinc with copper(II) sulfate solution.
Zn (s) + Cu2+ (aq) → Zn2+ (aq) + Cu (s). The sulfate ion (SO42-) is a spectator ion and is cancelled from both sides.
True or False?
In a balanced chemical equation, atoms can be created or destroyed as long as the overall mass is conserved.
False.
Atoms cannot be created or destroyed in a chemical reaction. The number of each type of atom must be identical on both sides of the equation.
When balancing combustion reactions of organic compounds, balance the .......... atoms first, then .......... and finally .......... .
When balancing combustion reactions of organic compounds, balance the carbon atoms first, then hydrogen and finally oxygen.
What is the difference between a word equation and a symbol equation?
A word equation uses only words to show reactants and products. A symbol equation uses chemical symbols and formulae to show the number and type of each atom involved in the reaction.
Define empirical formula.
An empirical formula is the simplest whole-number ratio of the elements present in one molecule or formula unit of a compound. For example, the empirical formula of ethanoic acid (C2H4O2) is CH2O.
How do you calculate the empirical formula from the masses of elements present in a compound?
Divide the mass of each element by its relative atomic mass to find moles. Then divide all values by the smallest to get the simplest whole-number ratio.
True or False?
Ionic compounds always have the same empirical and molecular formulae.
True.
Ionic compounds are represented by their simplest formula unit, so their empirical and molecular formulae are always the same.
To find the molecular formula from the empirical formula, divide the .......... of the molecular formula by the .......... of the empirical formula, then multiply each subscript by this number.
To find the molecular formula from the empirical formula, divide the relative formula mass of the molecular formula by the relative formula mass of the empirical formula, then multiply each subscript by this number.
Define molecular formula.
A molecular formula is the exact number of atoms of each element present in one molecule of a compound. For example, ethanoic acid has the molecular formula C2H4O2.
A compound contains 85.7% carbon and 14.3% hydrogen by mass. What is its empirical formula? (Ar: C = 12.0, H = 1.0)
Moles of C = 85.7/12.0 = 7.14. Moles of H = 14.3/1.0 = 14.3. Ratio = 1:2. Empirical formula = CH2.
True or False?
Organic molecules always have the same empirical and molecular formulae.
False.
Organic molecules often have different empirical and molecular formulae. For example, ethanoic acid has empirical formula CH2O and molecular formula C2H4O2.
A compound contains 10 g of hydrogen and 80 g of oxygen. The mole ratio of H to O is .......... to .......... giving the empirical formula .......... .
A compound contains 10 g of hydrogen and 80 g of oxygen. The mole ratio of H to O is 2 to 1 giving the empirical formula H2O.
The empirical formula of compound X is C4H10S and its relative formula mass is 180. What is the molecular formula of X? (Ar: C = 12, H = 1, S = 32)
Mr of empirical formula = (4x12) + (10x1) + 32 = 90. Ratio = 180/90 = 2. Molecular formula = C8H20S2.
Define water of crystallisation.
Water of crystallisation is water that is chemically incorporated into the crystal structure of a compound. It is shown in the formula by a dot, e.g. CuSO4•5H2O.
True or False?
A hydrated compound contains water of crystallisation as part of its structure.
True.
A hydrated compound incorporates water molecules into its crystal lattice. An anhydrous compound contains no water of crystallisation.
How can you convert a hydrated salt to its anhydrous form, and how is this reversed?
Heating the hydrated salt drives off the water of crystallisation to give the anhydrous compound. Adding water reverses the process, reforming the hydrated salt.
The formula of hydrated copper(II) sulfate is .......... and the formula of anhydrous copper(II) sulfate is .......... .
The formula of hydrated copper(II) sulfate is CuSO4•5H2O and the formula of anhydrous copper(II) sulfate is CuSO4.
Define anhydrous compound.
An anhydrous compound is a compound that contains no water of crystallisation. It is the form obtained when a hydrated salt is fully dried by heating.
Describe the experimental steps used to determine the degree of hydration (water of crystallisation) of a hydrated salt.
Measure the mass of the hydrated salt before heating. Heat until constant mass is achieved. Calculate the mass of water lost, then use mole ratios (dividing by respective Mr values) to find the number of water molecules.
True or False?
A compound can only be hydrated to one fixed degree.
False.
A compound can be hydrated to different degrees. For example, cobalt(II) chloride can form CoCl2•6H2O or CoCl2•2H2O.
10.0 g of CuSO4•xH2O is heated to give 5.59 g of CuSO4. The mass of water lost is .......... g. Dividing by the Mr of water (18.0) gives .......... mol H2O.
10.0 g of CuSO4•xH2O is heated to give 5.59 g of CuSO4. The mass of water lost is 4.41 g. Dividing by the Mr of water (18.0) gives 0.245 mol H2O.
How is a water of crystallisation calculation similar to an empirical formula calculation?
Both involve dividing masses by formula masses to find moles, then dividing by the smallest value to get the simplest ratio. In water of crystallisation calculations the salt and water replace the individual elements.
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