Week 3 - Stoichiometry Flashcards
Balance the equation AgNO3 + H2S → Ag2S + HNO3
There are two Ag on the right but only one on the left. Put 2 in front of AgNO3.
2 AgNO3 + H2S → Ag2S + HNO3
There are now two N on the left and one on the right. Put 2 in front of HNO3.
2 AgNO3 + H2S → Ag2S + 2 HNO3
Now check - there are two Ag on each side, two N on each side, six O on each side (2 × 3), two H on each side, and two H on each side.
Balance the equation C4H10 + O2 → CO2 + H2O
There are four C on the left but only one on the right. Put 4 in front of CO2.
C4H10 + O2 → 4 CO2 + H2O
There are ten H on the left and two on the right. Put 5 in front of H2O.
C4H10 + O2 → 4 CO2 + 5 H2O
There are two O on the left and 13 on the right (8 + 5). To have 13 O on the left, you would need 6.5 O2. However, we need to use whole numbers. Let’s double everything:
2 C4H10 + O2 → 8 CO2 + 10 H2O
Now we have 26 O on the right. Put 13 in front of the O2.
2 C4H10 + 13 O2 → 8 CO2 + 10 H2O
Now check - we have 8 C, 20 H, and 26 O on each side.
What are the symbols used in balanced equations?
Symbol Meaning
(s) Solid state (written after substance)
(l) Liquid state (written after substance)
(g) Gaseous state (written after substance)
(aq) Aqueous state (when substance is dissolved in water)
Δ Heat is added (when written above or below the arrow)
Symbols and meanings of chemical equations
Symbol Meaning
(s) Solid state (written after substance)
(l) Liquid state (written after substance)
(g) Gaseous state (written after substance)
(aq) Aqueous state (when substance is dissolved in water)
Δ Heat is added (when written above or below the arrow)
What is a decomposition reaction?
A single substance is decomposed, or broken down, to give two or more different substances.
2 HgO(s) → 2 Hg(l) + O2(g) CaCO3(s) → CaO(s) + CO2(g) 2 KClO3(s) → 2 KCl(s) + 3O2(g)
What is a single-displacement reaction?
One element reacts with a compound to replace one of the elements of that compound, yielding a different element and a different compound.
Zn(s) + 2 HCl(aq) → H2(g) + ZnCl2(aq) 2 Al(s) + 3 H2SO4(aq) → 3 H2(g) + Al2(SO4)3(aq)
What is a double-displacement reaction
Two compounds exchange partners with each other to produce two different compounds.
HCl(aq) + NaOH(aq) → NaCl(aq) + H2O(l)
CuO(s) + 2 HNO3(aq) → Cu(NO3)2(aq) + H2O(l)
Equation for percentage composition (formula) of molecules
% element = (number of atoms of the element) x (atomic mass of element) / formula mass of compound x 100
Avogadro’s number
A mole used to be defined as the amount of matter that contains as many objects as the number of atoms in exactly 12 g of pure 12C. 12 g of has 6.022 × 1023 of atoms.
In 2017, a new definition of mole was proposed, with IUPAC Recommendations, to state:
“One mole contains exactly 6.022 140 76 × 1023 elementary entities. This number is the fixed numerical value of the Avogadro constant, NA, when expressed in mol−1, and is called the Avogadro number (Pure and Applied Chemistry 2018, 90(1), p175–180).”
Molar Mass
The atomic mass of an element in grams contains Avogadro’s number of atoms and is defined as the molar mass of that element.
To determine molar mass, we change the units of the atomic mass (found in the periodic table) from atomic mass unit (amu) to grams.
For example, sulfur has atomic mass of 32.07 amu (for one sulfur atom), so 1 mol (6.022×1023 sulfur atoms) of sulfur has a molar mass of 32.07 g.
formula for Mole
Mole = Mass (g) / Molar Mass
How many magnesium atoms are contained in 5.00g Mg?
Molar mass of Mg = 24.31 g/mol, therefore,
Moles = Mass / Molar Mass = 5.00 g / 24.31 g/mol = 0.206 mol
Because 1 mole contains 6.022 × 1023 items (Mg atoms in this case), 0.206 mol has 0.206 mol × 6.022 × 1023 = 1.24 × 1023 Mg atoms
Define Stoichiometry
Measuring chemicals that go into, and come out of, and given reaction
How to find mass (equation)
Mass (g) = moles (mol) x molar mass (g/mol)
How many molecules in 40.03 mg of ammonium nitrate (NH4)NO3?
Referring to the above equations we first need to calculate the molar mass of ammonium nitrate, then calculate the number of moles this equated to and finally we can determine the number of molecules in that amount.
1
1
Calculate the molar mass of ammonium nitrate
2 N = 2 x 14.01 = 28.02 g/mol
4 H = 4 x 1.01 = 4.04 g/mol
3 O = 3 x 16.00 = 48.00 g/mol
Molar mass = = 80.06 g/mol
2
2
Calculate the number of moles of ammonium nitrate
Moles = 40.03 mg / 80.06 g/mol
Moles = 5.000 x 10-4 moles
3
3
Calculate the number of molecules of ammonium nitrate
Number of molecules = moles x Avogadro’s Number
Number of molecules = 5.000 x 10-4 x 6.022 x 1023
Number of molecules = 30.11 x 1019 molecules of ammonium nitrate
Difference between empirical and molecular formula
The empirical formula, or simplest formula, gives the smallest whole-number ratio of atoms present in a compound.
The molecular formula is the true formula, representing the total number of atoms or each element present in one molecule of a compound.
For example, acetylene (C2H2) is a common gas used in welding; Benzene (C6H6) is an important solvent used in synthesis of styrene and nylon. Both contain 92.3% C and 7.7% H. The smallest ratio of C and H for both compounds is 1:1. They have the same empirical formula but different molecular formulae.
What is a limiting reactant
For example, using 80 wheels and 50 frames to assemble bicycles, the number of bicycles that can be built from these parts is determined by the “limiting reactant” (the wheels in this case). There are enough frames for 50 bicycles, but only enough wheels for 40 bicycles. Once those wheels have been used, you can’t make any more bicycles.
Theoretical yield and formula
The quantities of the products we have been calculating from chemical equations represent the maximum yield (100%), called the theoretical yield, of product according to the reaction represented by the equation.
percent yield = actual yield/theoretical yield *100
(Molarity)
What is the molarity of a solution made by dissolving 5.85 g of sodium chloride (NaCl) in water to make 250. mL of solution?
Molarity (mol/L) = moles / volume (L)
Molar mass of NaCl= 58.45 g/mol
Moles of NaCl = 5.85 g / 58.45 g/mol = 0.100 mol
Molarity = 0.100 mol / 0.250 L = 0.400 M
(Molarity)
What is the molarity of a solution made by dissolving 2.00 g of potassium chlorate (KClO3) in enough water to make 150. mL of solution?
Molarity (mol/L) = moles / volume (L)
Molar mass of KClO3 = 122.6 g/mol
Moles of KClO3 = 2.00 g / 122.5 g/mol = 0.0163 mol
Molarity = 0.0163 mol / 0.150 L = 0.109 M
Dilution
As only water is added in the dilution process, the moles of solute before dilution are the same as the moles of solute after dilution. We can express this in the following equation:
MiVi = moles = MfVf or simply as MiVi = MfVf
Where the subscript i represents the initial and the subscript f the final molarity and volumes of a solution.
This is sometimes written as C1V1 = C2V2
Dilution question
What mass of lead chloride [PbCl2] could be obtained when 50 mL of a 0.200 M lead nitrate solution [Pb(NO3)2] is reacted with 250 mL of 0.100 M hydrochloric acid [HCl]?
Step 1 : Start with the balanced equation:
Pb(NO3)2 (aq) + 2 HCl (aq) → PbCl2 (s) + 2 HNO3 (aq)
Step 2 : Calculate the number of moles of lead nitrate and hydrochloric acid
Moles = Molarity (mol/L) x volume (L)
Lead Nitrate: 0.200 mol/L x 0.050 L = 0.01 mol
Hydrochloric acid: 0.100 mol/L x 0.250 L = 0.025 mol
Step 3 : Determine which is the limiting reagent by dividing by the reaction coefficient
Lead nitrate = 0.01 mol /1 = 0.01 mol
Hydrochloric acid = 0.025 mol/2 = 0.0125 mol
Lead nitrate is therefore the limiting reagent
Step 4 : Calculate the mass lead chloride which could be obtained.
0.01 moles lead nitrate would react to give 0.01 moles lead chloride (1:1 ratio)
Mass (g) = moles x molar mass (g/mol)
Mass lead chloride (g) = 0.01 mol x 278.1 g/mol = 2.78 g
