Module 2: Section 1- Atoms and Reactions

Question 1

What is mass number?

Answer 1

The total number of protons and neutrons in the nucleus.

Question 2

What is atomic number?

Answer 2

The number of protons in the nucleus of an atom. The number of protons is what identifies the element; all atoms of the same element has same number of protons. The periodic table is also arranged by atomic number.

Question 3

What is an ion?

Answer 3

An ion is a positively or negatively charged atom (due to an imbalance of protons and electrons) or covalently bonded group of atoms.

Question 4

What is an isotope?

Answer 4

Isotopes are atoms of the same element with a different number of neutrons (but same number of protons).
Isotopes of the same element will the same chemical properties as the configuration of electrons will be the same (if they are not ions). However, the chemical properties may differ slightly, as this is dependent on the mass of the atom.

Question 5

State some chemical properties.

Answer 5

Flammability, Solubility, Conductivity, Enthalpy, Oxidation, Toxicity, Acidity.

Question 6

State some physical properties.

Answer 6

Densities, Rate of diffusion, Melting and boiling point, Flexibility.

Question 7

What is relative atomic mass?

Answer 7

The weighted mean mass of an atom of an element compared to 1/12th the mass of an atom of carbon-12.

Question 8

What is relative molecular mass?

Answer 8

The weighted mean mass of a molecule compared with 1/12th the mass of an atom of carbon-12.

Question 9

What is relative isotopic mass?

Answer 9

The mass of an atom of an isotope, compared with 1/12th the mass of an atom of carbon-12.

Question 10

How to calculate relative atomic mass from isotopic abundances?

Answer 10

1. Multiply each relative isotopic mass by isotopic abundance. Add all of these together.
2. Divide answer by 100.

Question 11

On a mass spectra (produced my mass spectrometry) for isotopes of an element, what is found on the x and y axis?
What can be calculated from the information of a mass spectra for isotopes?

Answer 11

On the x axis, we find the mass/ charge ratio, in other words, the relative isotopic mass. Then, the y axis, tells us the abundance (as a percentage) of each isotope.
As we have the relative isotopic mass and respective abundances for each isotope, we can calculate relative atomic mass for that element.

Question 12

Find the relative atomic mass of boron given that 20% of the boron atoms found on Earth has a relative isotopic mass of 10.0, while 80% have a relative isotopic mass of 11.0.

Answer 12

(20 x 10) + (80 x 11) = 1080
1080 / 100= 10.8

Question 13

What is Avogadro’s Constant a measure of ?
State Avogadro’s Constant.

Answer 13

The number of particles (e.g. atoms, molecules) in a mole.
Avogadro’s Constant = 6.022 x 10^23

Question 14

How can we calculate the moles of a compound, given we have the mass?

Answer 14

Moles= Mass / Molar mass

Question 15

How many moles of chlorine molecules are present in 71.0g of chlorine gas?

Answer 15

Moles= Mass / Molar mass
Mass = 71.0g

As it is a molecule, it is Cl 2, rather than just Cl.
Molar mass of Cl 2 = 35.5 x 2 = 71

Moles= 71 / 71 = 1 mol

Question 16

At room temperature and pressure, 1 mole of any gas has the same volume.
What is room temperature and pressure?
What is the volume each mole takes up?

Answer 16

Room temperature= 25 degrees Celsius (298K)
Room pressure= 1 atm (101kPa)
Volume taken by each mole= 24 dm^3

Question 17

What is the formula to calculate the volume of a gas at room temperature and pressure?

Answer 17

Volume of gas= moles x molar gas volume
Where molar gas volume is a constant, which is 24.

Question 18

What equation is used to calculate the volume of gases that are not at room temperature and pressure?
State the units of each component of equation.

Answer 18

pV= nRT
p = pressure (Pa)
V = volume (m^3)
n = number of moles
R = 8.314 (this is a constant)
T = temperature (K)

Question 19

How to go from cm^3 to m^3?
How to go from dm^3 to m^3?
How to go from dm^3 to cm^3?

Answer 19

Go from cm^3 to m^3= divide by 1000000
Go from dm^3 to m^3= divide by 1000
Go from dm^3 to cm^3= multiply by 1000

Question 20

How to go from kPa to Pa?

Answer 20

kPa to Pa= multiply by 1000

Question 21

How to go from degrees Celsius to kelvin?

Answer 21

Value in Kelvin= Value in Degree Celsius + 273

Question 22

What is the difference between empirical and molecular formula?

Answer 22

Empirical formula is the smallest whole number ratio of atoms in a compound. Whereas, molecular formula is actual number of atoms of each element in a compound.

Question 23

A molecule has empirical formula C4 H3 O2, and molecular mass of 166 g/mol. Work out its molecular formula.

Answer 23

Empirical mass= 4(12) + 3(1) + 2(16) = 83 g/mol
166 / 83 = 2 empirical units
Therefore do empirical formula x 2, which is
C8 H6 O4

Question 24

When a hydrocarbon is burnt in excess oxygen, 4.4g of carbon dioxide and 1.8g of water is made. What is the empirical formula of the hydrocarbon?

Answer 24

moles = mass / Molar mass
mol of carbon dioxide = 4.4 / 44 = 0.1mol

1 mol of carbon dioxide contains 1 mol of carbon atoms, so 0.1 mol of carbon dioxide contains 0.1 mol of carbon atoms.

mol of water = 1.8 / 18 = 0.1 mol

1 mol of water, contains 2 mol of hydrogen atoms, so 0.1 mol of water contains 0.2 mol hydrogen atoms.

Ratio of carbon atoms : hydrogen atoms
0.1 : 0.2
1 : 2
So empirical formula of hydrocarbon is CH2.

Question 25

A compound is found to have percentage composition of 56.5% potassium, 8.70% carbon and 34.8% oxygen by mass. Calculate its empirical formula.

Answer 25

Assuming there is 100g of the compound, the percentage composition of each element would equate to the element’s mass.

1. First find the mole of each element by doing their mass (respective percentage composition) divided by their relative atomic mass.
   (Like in the formula: mol = mass / molar mass)

mol of potassium = 56 / 39.1 = 1.45 mol
mol of carbon = 8.70 / 12 = 0.725 mol
mol of oxygen = 34.8 / 16 = 2.18 mol

2. Then divide each answer obtained, by the smallest value of moles, here the smallest value is 0.725.

For potassium: 1.45 / 0.725 = 2.00
For carbon= 0.725 / 0.725 = 1.00
For oxygen= 2.18 / 0.725 = 3.01

These numbers equate to the moles of each element in the empirical formula. Hence, the empirical formula is:
K2 CO3

Question 26

What is an ionic equation?

Answer 26

An ionic equation only shows the reacting ions of the reaction.

Question 27

What are the steps to forming an ionic equation from a fully balanced equation of the whole reaction?

Answer 27

1. Take the full balanced equation and separate each component into their ions. (Keep covalent molecules like water the same).
2. Now the equation is split into the individual ions, cancel out any spectator ions.
3. Balance the ionic equation so that the charges are equal on both sides.

Question 28

What is hydrated and anhydrous compounds?

Answer 28

Hydrated compounds contains water.
Anhydrous compounds do not contain water.

Question 29

What is water of crystallisation?

Answer 29

Water of crystallisation is the number of moles of water that forms part of a hydrated compound.

Question 30

Heating 3.210 g of hydrated magnesium sulfate Mg SO4. xH20, forms 1.567g of anhydrous magnesium sulfate. Find the value of x and write the formula of the hydrated salt.

Answer 30

1. First find the number of moles of water lost.
   Mass of water lost= 3.210 - 1.567 = 1.643 g
   Mol of water lost = 1.643 / 18 = 0.09127 mol
2. Find number of moles of anhydrous salt.
   Mol of anhydrous salt= 1.567 / 120.4 = 0.01301 mol
3. Form a ratio of anhydrous salt : water lost
   0.01301 : 0.09127
   (0.01301 / 0.01301) : (0.09127 / 0.01301)
   1 : 7.015 mol of water

Hence formula of hydrated salt is MgSO4. 7H2O
where x = 7