Topic: Moles

Reacting masses quizzes

Reacting masses. Year9/10 reactions. Chemical equations given (3 points per correct answer):

Reacting masses. Year9/10 reactions. Word equations given (4 points per correct answer):

Reacting masses. Year9/10 reactions. No equations given (5 points per correct answer):

Reacting masses. All reactions. Chemical equations given (3 points per correct answer):

Reacting masses. All reactions. Word equations given (4 points per correct answer):

Reacting masses. All reactions. No equations given (5 points per correct answer):

2018-11-24T15:53:43+00:00Categories: Uncategorized|Tags: , |

1:25 write word equations and balanced chemical equations (including state symbols): for reactions studied in this specification and for unfamiliar reactions where suitable information is provided

Example:

Sodium (Na) reacts with water (H2O) to produce a solution of sodium hydroxide (NaOH) and hydrogen gas (H2).

Word equation:

     sodium + water –> sodium hydroxide + hydrogen

Writing the chemical equation

A chemical equation represents what happens in terms of atoms in a chemical reaction.

Step 1: To write a chemical equation we need to know the chemical formulae of the substances.

     Na + H2O –> NaOH + H2

Step 2: The next step is to balance the equation: write a large number before each compound so the number of atoms of each element on the left hand side (reactants) matches the number on the right (products). This large number is the amount of each compound or element.

During this balancing stage the actual formulas for each compound must not be changed. Only the number of each compound changes.

     2Na + 2H2O –> 2NaOH + H2

If asked for an equation, the chemical equation must be given.

 

State symbols are used to show what physical state the reactants and products are in.

State symbolsPhysical state
(s)Solid
(l)Liquid
(g)Gas
(aq)Aqueous solution (dissolved in water)

Example:

A solid piece of sodium (Na) reacts with water (H2O) to produce a solution of sodium hydroxide (NaOH) and hydrogen gas (H2).

     2Na(s) + 2H2O(l) –> 2NaOH(aq) + H2(g)

1:26 calculate relative formula masses (including relative molecular masses) (Mᵣ) from relative atomic masses (Aᵣ)

Relative formula mass (Mr) is mass of a molecule or compound (on a scale compared to carbon-12).

It is calculated by adding up the relative atomic masses (Ar) of all the atoms present in the formula.

Example:

The relative formula mass (Mr) for water (H2O) is 18.

Water                     = H2O

Atoms present      = (2 x H) + (1 x O)

Mr                           = (2 x 1) + (1 x 16) = 18

1:27 know that the mole (mol) is the unit for the amount of a substance

In Chemistry, the mole is a measure of amount of substance (n).

The abbreviation for mole is mol.

The mass of 1 mole of a substance is the relative formula mass (Mr) of the substance in grams.

Example:

The Mr of water is 18.

Therefore the mass of 1 mol of water equals 18 g.

1:28 understand how to carry out calculations involving amount of substance, relative atomic mass (Aᵣ) and relative formula mass (Mᵣ)

The following formula allows for the interconversion between a mass in grams and a number of moles for a given substance:

    \[Amount(mol) = \frac{mass(g)}{M_r} \]

Example 1:

Calculate the amount, in moles, of 8.8 g of carbon dioxide (CO2).

Step 1: Calculate the relative formula mass (Mr) of carbon dioxide (CO2).

    \[Carbon\,dioxide = CO_2 \]

    \[Atoms\,present = (1 \times C) + (2 \times O) \]

    \[M_r = (1 \times 12) + (2 \times 16) = 44 \]

Step 2: Use the formula to calculate the amount in moles.

    \[Amount = \frac{mass(g)}{M_r} \]

    \[Amount = \frac{8.8}{44} \]

    \[Amount = 0.2 mol \]

Example 2:

Calculate the mass of 2 mol of copper(II) sulfate (CuSO4).

Step 1: Calculate the relative formula mass (Mr) of copper(II) sulfate (CuSO4).

    \[copper(II)\,sulfate = CuSO_4 \]

    \[Atoms\,present = (1 \times Cu) + (1 \times S) + (4 \times O) \]

    \[M_r = (1 \times 63.5) + (1 \times 32) + (4 \times 16) = 159.5 \]

Step 2: Rearrange the formula to calculate the mass.

    \[mass = amount \times M_r \]

    \[mass = 2 \times 159.5 \]

    \[mass = 319g\]

1:29 calculate reacting masses using experimental data and chemical equations

Example: When calcium carbonate (CaCO3) is heated calcium oxide is produced. You can use reacting mass calculations to calculate the mass of calcium oxide produced when heating 25 g of calcium carbonate.

     CaCO3     –>         CaO      +      CO2

Step 1: Calculate the amount, in moles, of 25 g of calcium carbonate (CaCO3)

    \[M_r(CaCO_3)=(1 \times 40) + (1\times12) + (3 \times 16) \]

    \[M_r(CaCO_3)= 100\]

    \[Amount = \frac{mass(g)}{M_r} \]

    \[Amount = \frac{25}{100} \]

    \[Amount = 0.25 mol \]

Step 2: Deduce the amount, in moles, of CaO produced from 0.25 mol of CaCO3.

This step involves using the ratio of CaCO3 to CaO from the chemical equation.

     CaCO3     –>         CaO      +      CO2

From the equation you can see that the ratio of CaCO3 to CaO is 1:1.

Therefore if you have 0.25 mol of CaCO3 this will produced 0.25 mol of CaO.

Step 3: Calculate the mass of 0.25 mol of CaO.

    \[M_r(CaO)=(1 \times 40) + (1 \times 16) \]

    \[M_r(CaO)= 56\]

    \[Mass(CaO)=amount \times M_r \]

    \[Mass(CaO)= 0.25 \times 56\]

    \[Mass(CaO)= 14g\]

 

A simple format for laying out this method can be used.

Example: What mass of ammonia (NH3) is formed when 7 g of nitrogen (N2) is combined with hydrogen (H2).

     

 

1:30 calculate percentage yield

Yield is how much product you get from a chemical reaction.

The theoretical yield is the amount of product that you would expect to get. This is calculated using reacting mass calculations.

In most chemical reactions, however, you rarely achieved your theoretical yield.

For example, in the following reaction:

     CaCO3     –>            CaO         +           CO2

You might expect to achieve a theoretical yield of 56 g of CaO from 100 g of CaCO3.

However, what if the actual yield is only 48 g of CaO.

By using the following formula, the % yield can be calculated:

    \[yield= \frac{actual\,amount\,of\,product}{theoretical\,amount\,of\,product} \]

    \[yield= \frac{48}{56} \]

    \[yield=0.86 \]

    \[\% yield=86\% \]

1:35 (Triple only) understand how to carry out calculations involving gas volumes and the molar volume of a gas (24dm³ and 24,000cm³ at room temperature and pressure (rtp))

The molar volume of a gas is the volume that one mole of any gas will occupy.

1 mole of gas, at room temperature and pressure (rtp), will always occupy 24 dm3 or 24,000 cm3.

Note: 1 dm3 = 1000 cm3

The following formulae allows for the interconversion between a volume in dm3 or cm3 and a number of moles for a given gas:

    \[Amount(mol) = \frac{volume(dm^3)}{24} \]

    or

    \[Amount(mol) = \frac{volume(cm^3)}{24000} \]

 

Example 1:

Calculate the amount, in moles, of 12 dm3 of carbon dioxide (CO2).

    \[Amount = \frac{volume(dm^3)}{24} \]

    \[Amount = \frac{12}{24} \]

    \[Amount = 0.5mol \]

 

Example 2:

Calculate the volume at rtp in cubic centimetres (cm3), of 3 mol of oxygen, (O2).

    \[Volume = amount \times 24000 \]

    \[Volume = 3 \times 24000 \]

    \[Volume = 72000cm^3 \]

 

Select a set of flashcards to study:

     Terminology

     Skills and equipment

     Remove Flashcards

Section 1: Principles of chemistry

      a) States of matter

      b) Atoms

      c) Atomic structure

     d) Relative formula masses and molar volumes of gases

     e) Chemical formulae and chemical equations

     f) Ionic compounds

     g) Covalent substances

     h) Metallic crystals

     i) Electrolysis

 Section 2: Chemistry of the elements

     a) The Periodic Table

     b) Group 1 elements: lithium, sodium and potassium

     c) Group 7 elements: chlorine, bromine and iodine

     d) Oxygen and oxides

     e) Hydrogen and water

     f) Reactivity series

     g) Tests for ions and gases

Section 3: Organic chemistry

     a) Introduction

     b) Alkanes

     c) Alkenes

     d) Ethanol

Section 4: Physical chemistry

     a) Acids, alkalis and salts

     b) Energetics

     c) Rates of reaction

     d) Equilibria

Section 5: Chemistry in industry

     a) Extraction and uses of metals

     b) Crude oil

     c) Synthetic polymers

     d) The industrial manufacture of chemicals