Elements, Compounds, Chemical Formula Writing and Naming, Moles, & Concentration
I. Elements & Compounds
a. The universe is made up of energy and matter
b. Matter is made up of building blocks called elements (~100 or so)
c. Each element is made of one type of atom
d. The elements combine together to form compounds
e. Atoms combine together to form molecules (millions and millions of different kinds)
f. Do “Elements & Compounds Review & Practice” WS
II. Chemical formulas – Writing
a. A chemical formula consists of symbols of the elements and subscripts (small numbers), which indicate the number of each element present in the compound.
i. Simple rules in writing Chemical Formulas:
1. Represent each kind of element in a compound with the correct symbol for that element
2. Use subscripts to indicate the number of atoms of each type in the compound. If there is only one atom of a particular element, no subscript is used
3. Write the symbol for the metallic element first.
4. Use the ion charges to balance out the charges. (All molecules are neutral; net charge = 0)
5. Special case for polyatomic ions: keep the whole ployatomic group together and use brackets if necessary.
1. Sodium Iodide NaI
2. magnesium oxide MgO
3. zinc chloride ZnCl2
4. potassium nitrate KNO3
5. Aluminum sulphate Al2(SO4)3
6. Lead (IV) oxide PbO2 (Why the IV?)
III. Chemical Formula – Naming
a. By looking at the above examples and your own knowledged gained in previous courses you should be able to develop a “feel” for naming chemical formulas.
b. Some simple rules:
i. Name metallic element first – unchanged
ii. Name non-metallic element second – change ending to –ide
iii. If polyatomic ion is present use its name as given
iv. If the metal has more than one different charge i.e. Iron comes in two forms: Fe2+ and Fe3+ then use a roman numeral to state which one is present.
1. Na2O sodium oxide
2. MgF2 magnesium fluoride
3. Co2O3 cobalt (III) oxide
4. PbSO4 lead (II) sulphate
5. (NH4)3PO4 ammonium phosphate
vi. Review Worksheet
IV. Balancing Chemical Reaction Equations
a. Word Equation – states the reactant compounds that are reacting to produce product compounds.
i. Sodium + chlorine à sodium chloride
b. Translating this reaction means to re-write the reaction using the correct chemical formulas for each reactant and product.
i. Na + Cl2 à NaCl
c. Balancing the chemical reaction means to ensure that each side of the equation has the same number and kind of atom. Is this chemical equation balanced?
i. To balance chemical reaction equations we use coefficients in front of the chemical formulas. DO NOT ALTER THE SUBSCIPTS!
1. 2 Na + 1Cl2 à 2 NaCl (Note: 1’s can be left out)
ii. Check to see if you are balanced by counting up the atoms on both sides
d. Please note: In chemistry 12 a lot of our discussions and areas of studies depend on being able to create and balance a chemical reaction equation. This may depend on writing a correct chemical formula. In order to do well in Chemistry 12, students have to be able to write chemical formulas and balance chemical reaction equations easily.
e. Review worksheet #18 & 19
V. Moles (Furry Creatures?, Skin blemish?, or maybe… a chemistry term!)
a. Moles are very important in chemistry, but remember it just a number! (Dozen = 12, 1 gross = 144, 1 case = 12 etc.)
1 mole = 6.02 x 1023
This number was chosen so that when you have this many atoms of any element, the mass of all the atoms would equal the atomic mass given on the periodic table.
i.e. carbon 1 mole = 6.02 x 1023 = 12.00 g
oxygen 1 mole = 6.02 x 1023 = 16.00 g
b. Calculating the molar mass of a compound:
i. Molar Mass = mass of 6.02 x 1023 molecules of that compound
ii. To determine the mass of this many molecules we take the mass of one proton and then multiply it by … just kidding!!!
1. Molar Mass of a Molecule:
a. Find the molar mass of each element in the compound according to the periodic table.
b. Multiple each mass by the number of each element in the compound.
c. Add up all the individual masses:
i. Example: NaNO3
Na 23.0 g x 1 = 23.0 g
N 14.0 g x 1 = 14.0 g
O 16.0 g x 3 = 48.0 g
Molar Mass of NaNO3 = 85.0 g/mole
d. Now Try These:
i. NaBr _____________
ii. MnCl2 _____________
iii. Al2O3 _____________
iv. (NH4)3PO4 _____________
v. CoCl2*5H20 _____________
c. Mole Conversions
i. In Chemistry 12 converting from moles to mass and back again occurs all the time. I know that this is taught in Chemistry 11. I will not spend a great deal of time on this topic but here is a real quick review:
ii. Moles to mass:
Mass (g) = Molar Mass x # of moles
i.e. What is the mass of 0.065 moles of NaCl?
Mass = 0.065 moles x 58.5 g/mole = 3.8 g of NaCl
iii. Mass to moles:
Moles = Mass / Molar Mass
i.e. How many moles are there in 12.0 g of MgF2?
First find the molar mass of MgF2: 62.3 g/mole
Then: # moles = 12.0 g / 62.3 g/mole = 0.193 moles MgF2
VI. Reaction and Mole Ratios
a. Chemical reaction Equations were not created to make chemistry students live difficult!
b. They actually serve an important purpose.
c. Remember this pervious reaction:
2 Na + 1Cl2 à 2 NaCl
The coefficients state the mole ratio for the reaction between Na and Cl2.
2 Na + 1Cl2 à 2 NaCl
For example: This reaction means:
2 moles of Na will combine with 1 mole of Cl2 to form 2 moles of NaCl
4 moles of Na will combine with 2 mole of Cl2 to form 4 moles of NaCl
0.01 moles of Na will combine with 0.005 mole of Cl2 to form 0.01 moles of NaCl
2 Al(s) + 3 FeCO3(aq) à Al2(CO3)3(aq) + 3 Fe(s)
How many grams of iron (II) carbonate would be needed to completely react with 108 g of aluminum?
Step 1: Convert grams of Al to moles of Al
Step 2: Determine the mole ratio between Al and FeCO3.
Step 3: Find the number of moles of FeCO3 from the moles of Al and the mole ratio.
Step 4: Convert moles of FeCO3 back to gram of FeCO3.
VII. Concentration Calculations:
a. Once again this topic was covered in Chemistry 11. I will only give you a quick over view of it here. All of Chemistry 12 is solution chemistry, which means that calculating concentrations and using concentrations to solve problems is an important concept to know.
b. Concentration = moles / volume
What is the concentration of a solution made when 19.0 g of Mg(NO3)2 is dissolved in a total volume of 250 mL of water?
i. Find the molar mass of Mg(NO3)2:
Mg(NO3)2 = 148.3 g/mole
ii. Find the number of moles from this molar mass and the mass given:
# moles = 19.0g / 148.3 g/mole = 0.12811 moles
iii. Calculate the concentration of the solution:
C = n / V = 0.12811 moles / 0.250 L
= 0.512 M Mg(NO3)2 Solution
d. Now try this one:
A student needs to prepare 125 mL solution of a 0.300 M lead (II) nitrate solution. How many grams of the lead (II) nitrate does he need?