Moles Balanced chemical equations Molar ratios Mass Composition Empirical and Molecular Mass Predicting Quantities Equations
Micro World atoms & molecules Macro World grams Atomic mass is the mass of an atom in atomic mass units (amu) By definition: 1 atom 12 C weighs 12 amu On this scale 1 H = 1.008 amu 16 O = 16.00 amu 3.1
The mole (mol) is the amount of a substance that contains as many elementary entities as there are atoms in exactly 12.00 grams of 12 C 1 mol = N A = 6.0221367 x 10 23 Avogadro s number (N A ) 3.2
Molar mass is the mass of 1 mole of Na atoms Pb atoms Kr atoms Li atoms in grams 1 mole 12 C atoms = 6.022 x 10 23 atoms = 12.00 g 1 mole 12 C atoms = 12.00 g 12 C 1 mole lithium atoms = 6.941 g of Li For any element atomic mass (amu) = molar mass (grams) 3.2
Species H Quantity 1 mole Number of H atoms 6.022 x 10 23
Species H 2 Quantity 1 mole Number of H 2 molecules 6.022 x 10 23
Species Na Quantity 1 mole Number of Na atoms 6.022 x 10 23
Species C 6 H 6 Quantity 1 mole Number of C 6 H 6 molecules 6.022 x 10 23
1 mol of atoms = 6.022 x 10 23 atoms 1 mol of molecules = 6.022 x 10 23 molecules 1 mol of ions = 6.022 x 10 23 ions
Converting Units Unit Factor Label Method I. Write Unit of Answer II. Write Starting Quantity from Problem III. Add appropriate unit factor(s) to cancel out starting quantity and put in unit of answer.
Understanding Molar Mass How many atoms are in 0.551 g of potassium (K)? 1 mol K = 39.10 g K 1 mol K = 6.022 x 10 23 atoms K 0.551 g K x 1 mol K 39.10 g K x 6.022 x 1023 atoms K 1 mol K = 8.49 x 10 21 atoms K 3.2
How many moles of iron does 25.0 g of iron represent? Atomic mass iron = 55.85 Conversion sequence: grams Fe moles Fe Set up the calculation using a conversion factor between moles and grams. (grams Fe) 1 mol Fe 55.85 g Fe (25.0 g Fe) 1 mol Fe = 55.85 g Fe 0.448 mol Fe
What is the mass of 3.01 x 10 23 atoms of sodium (Na)? Molar mass Na = 22.99 g Conversion sequence: atoms Na grams Na Set up the calculation using a conversion factor between grams and atoms. 22.99 g Na (atoms Na) 6.022 x 10 23 atoms Na 22.99 g Na = 11.5 g Na 23 (3.01 x 10 atoms Na) 6.022 x 10 23 atoms Na
Molecular mass (or molecular weight) is the sum of the atomic masses (in amu) in a molecule. SO 2 1S 2O SO 2 32.07 amu + 2 x 16.00 amu 64.07 amu For any molecule molecular mass (amu) = molar mass (grams) 1 molecule SO 2 = 64.07 amu 1 mole SO 2 = 64.07 g SO 2 3.3
Calculate the molar mass of C 2 H 6 O. 2 C = 2(12.01 g) = 24.02 g 6 H = 6(1.01 g) = 6.06 g 1 O = 1(16.00 g) = 16.00 g 46.08 g
Calculate the molar mass of LiClO 4. 1 Li = 1(6.94 g) = 6.94 g 1 Cl = 1(35.45 g) = 35.45 g 4 O = 4(16.00 g) = 64.00 g 106.39 g
Calculate the molar mass of (NH 4 ) 3 PO 4. 3 N = 3(14.01 g) = 42.03 g 12 H = 12(1.01 g) = 12.12 g 1 P = 1(30.97 g) = 30.97 g 4 O = 4(16.00 g) = 64.00 g 149.12 g
In dealing with diatomic elements (H 2, O 2, N 2, F 2, Cl 2, Br 2, and I 2 ), distinguish between one mole of atoms and one mole of molecules.
Calculate the molar mass of 1 mole of H atoms. 1 H = 1(1.01 g) = 1.01 g Calculate the molar mass of 1 mole of H 2 molecules. 2 H = 2(1.01 g) = 2.02 g
How many grams of (NH 4 ) 3 PO 4 are contained in 2.52 moles of (NH 4 ) 3 PO 4? The molar mass of (NH 4 ) 3 PO 4 is 149.12 g. Conversion sequence: moles (NH 4 ) 3 PO 4 grams (NH 4 ) 3 PO 4 Use the conversion factor: 149.12 grams (NH ) PO 1 mole (NH ) PO 4 3 4 4 3 4 149.12 g (NH 4) 3PO4 (2.52 mol (NH 4) 3PO 4)) 1 mol (NH 4) 3PO4 = 376g (NH 4) 3PO4
56.04 g of N 2 contains how many N 2 molecules? The molar mass of N 2 is 28.02 g. Conversion sequence: g N 2 moles N 2 molecules N 2 Use the conversion factors 1 mol N 2 28.02 g N 1 mol N 2 (56.04 g N 2) 28.02 g N 2 23 6.022 x 10 molecules N2 2 1 mol N 2 23 6.022 x 10 molecules N2 1 mol N 24 = 1.204 x 10 molecules N 2 2
56.04 g of N 2 contains how many N 2 atoms? The molar mass of N 2 is 28.02 g. Conversion sequence: g N 2 moles N 2 molecules N 2 atoms N Use the conversion factors 23 1 mol N 2 6.022 x 10 molecules N 2 atoms N 2 28.02 g N2 1 mol N 1 molecule N2 1 mol N 2 (56.04 g N 2) 28.02 g N 2 2 23 6.022 x 10 molecules N2 1 mol N 2 2 atoms N 1 molecule N 2 24 = 1.204 x 10 molecules N 2
If the formula of a compound is known, a two-step process is needed to calculate the percent composition. Step 1 Calculate the molar mass of the formula. Step 2 Divide the total mass of each element in the formula by the molar mass and multiply by 100.
total mass of the element molar mass x 100 = percent of the element
Calculate the percent composition of hydrosulfuric acid H 2 S. Step 1 Calculate the molar mass of H 2 S. 2 H = 2 x 1.01g = 2.02 g 1 S = 1 x 32.07 g = 32.07 g 34.09 g
Calculate the percent composition of hydrosulfuric acid H 2 S. Step 2 Divide the mass of each element by the molar mass and multiply by 100. 2.02 g H H: (100) = 5.93% 34.09 g 32.07g S S: (100) = 94.07% 34.09g S 94.07% H 5.93%
Percent Composition From Experimental Data
Percent composition can be calculated from experimental data without knowing the composition of the compound. Step 1 Calculate the mass of the compound formed. Step 2 Divide the mass of each element by the total mass of the compound and multiply by 100.
A compound containing nitrogen and oxygen is found to contain 1.52 g of nitrogen and 3.47 g of oxygen. Determine its percent composition. Step 1 Calculate the total mass of the compound 1.52 g N 3.47 g O 4.99 g = total mass of product
Calculate the percent composition of hydrosulfuric acid H 2 S. Step 2 Divide the mass of each element by the total mass of the compound formed. 1.52 g N (100) = 30.5% 4.99 g O N 69.5% 30.5% 3.47g O (100) = 69.5% 4.99g
Percent composition of an element in a compound = n x molar mass of element molar mass of compound x 100% n is the number of moles of the element in 1 mole of the compound C 2 H 6 O 2 x (12.01 g) %C = 46.07 g 6 x (1.008 g) %H = 46.07 g 1 x (16.00 g) %O = 46.07 g x 100% = 52.14% x 100% = 13.13% x 100% = 34.73% 52.14% + 13.13% + 34.73% = 100.0% 3.5
Empirical Formula versus Molecular Formula
The empirical formula or simplest formula gives the smallest whole-number ratio of the atoms present in a compound. The empirical formula gives the relative number of atoms of each element present in the compound.
The molecular formula is the true formula of a compound. The molecular formula represents the total number of atoms of each element present in one molecule of a compound.
Molecular Formula C 2 H 4 Empirical Formula CH 2 Smallest Whole Number Ratio C:H 1:2
Molecular Formula C 6 H 6 Empirical Formula CH Smallest Whole Number Ratio C:H 1:1
Molecular Formula H 2 O 2 Empirical Formula HO Smallest Whole Number Ratio H:O 1:1
Two compounds can have identical empirical formulas and different molecular formulas.
Calculating Empirical Formulas
Step 1 Assume a definite starting quantity (usually 100.0 g) of the compound, if not given, and express the mass of each element in grams. Step 2 Convert the grams of each element into moles of each element using each element s molar mass.
Step 3 Divide the moles of atoms of each element by the moles of atoms of the element that had the smallest value If the numbers obtained are whole numbers, use them as subscripts and write the empirical formula. If the numbers obtained are not whole numbers, go on to step 4.
Step 4 Multiply the values obtained in step 3 by the smallest numbers that will convert them to whole numbers Use these whole numbers as the subscripts in the empirical formula. FeO 1.5 Fe 1 x 2 O 1.5 x 2 Fe 2 O 3
The results of calculations may differ from a whole number. If they differ ±0.1 round off to the next nearest whole number. 2.9 3 Deviations greater than 0.1 unit from a whole number usually mean that the calculated ratios have to be multiplied by a whole number.
A gaseous compound containing carbon and hydrogen was analyzed and found to consist of 83.65% carbon by mass. Determine the empirical formula of the compound. 1.Obtain the mass of each element present (in grams). Assume you have 100 g of the compound. 83.65% C = 83.65 g C (100.00 83.65) 16.35% H = 16.35 g H
2. Determine the number of moles of each type of atom present.
3. Divide the number of moles of each element by the smallest number of moles to convert the smallest number to 1. If all of the numbers so obtained are integers (whole numbers), these are the subscripts in the empirical formula. If one or more of these numbers are not integers, go on to step 4.
4. Multiply the numbers you derived in step 3 by the smallest integer that will convert all of them to whole numbers. This set of whole numbers represents the subscripts in the empirical formula.
The balanced chemical equation represents the ratio Of one species to any others in the equation: NaOH (aq) + HCl (aq) NaCl (aq) + H 2 O (l) 1mole of NaOH produces 1mole of NaCl 1 mol NaCl 1mol NaOH 2H 2 O 2 (l) 2 H 2 O (l) + O 2 (g) 2 moles of H 2 O 2 (l) produces 1 mole of O 2 gas 1 mol O 2 2 mol H 2 O 2 (l)
Mole composition gives ratio of each element In a compound 0.345 moles of Al 2 (CO 3 ) 3 has the following moles of Al in the compound: Mol ratio of Al in Al 2 (CO 3 ) 3 : 0.345 mol Al 2 (CO 3 ) 3 X 2 mol Al = 0.690 mol Al 1 mol Al 2 (CO 3 ) 3 Moles of compound X mol ratio (element/molecule) = mol of element
Mass of Aluminum is obtained from the molar mass: 0.690 mol Al X 26.98 g = 18.62 g Al mol Al Mol X grams/mol = mass The mass of any substance can be determined if The molar mass is known. The molar mass comes From the periodic table.
233.984 g/mol X 0.345 mol = 80.724 g Al 2 (CO 3 ) 3 Al 2 (CO 3 ) 3 continued: C: 0.345 mol X 3 mol C = 1.04 mol C X 12.011g/mol 1 mol Al 2 (CO 3 ) 3 = 12.43 g C O: 0.345 mol X 9 mol O = 3.11 mol O X 15.999g/mol 1 mol Al 2 (CO 3 ) 3 = 49.677 g O Formula Weight converted to mass from moles: 2 mol X 26.98g/mol (Al) =53.96g 3mol X 12.011g/mol (C) =36.033g 9 mol X 15.999g/mol (O) = 143.991 g total molar mass 233.984 g in 1 mol Al 2 (CO 3 ) 3
The empirical formula for a compound is the Smallest whole number ratio between the Elements that make up the compound. This is not necessarily the actual formula, especially For non-ionic compounds such as organic compounds. To determine the empirical formula, we must know The percent composition of each element in a compound The percent composition is a MASS PERCENT and is Determined for 100 grams of the substance. This has The effect of converting the percent of each element Into a RATIO!
Determination of Empirical Formula from Mass % 1. Assume 100 grams of material 2. Determine moles of each element in compound 3. Divide all subscripts in empirical formula by lowest number because the number of atoms must be an integer
A compound is discovered to be 63.6% nitrogen And 36.4% oxygen. What is the empirical formula? Convert to mass: 63.6% N = 63.6 g N 100 g Compound 36.4% O= 36.4 g O 100 g compound II. Convert to moles: 63.6 g N X 1 mol N = 4.54 mol N 14.0 g N 36.4 g O X 1 mol O = 2.28 mol O 15.999 g O The ratio of N to O atoms will be the same as the Ratio of N to O moles: N: 4.54 = 1.99 O: 2.28 = 1.00 2.28 2.28 III Divide all mols by the SMALLEST number of mols Empirical Formula= N 2 O
For organic compounds and some non-organic Compounds, the molecular formula is the actual Ratio of elements. This is a multiple of the empirical Formula. We can discover the molecular formula from the Empirical one if we determine the molar mass from An experiment. We simply compare the experimental Molar mass with the empirical one.
Determination of Molecular Formula from Empirical Formula Only Empirical Formula can be Determined by Mass Percent Multiplier = MoleculeMolarMass EmpiricalMolarMass
A phosphorous compound has an empirical formula of P 2 O 5. It has a molar mass of 284 g/mol, determined in the laboratory. The empirical formula weight is: (2 x 30.974 g + 5 x 15.999 g)/mol= 141.943 g/mol Compare the masses in the molar ratio: 284 g/mol = 2; therefore:(p 2 O 5 ) 2 is the molecular 141.943g/mol Formula: P 4 O 10 Molar Mass from Experiment = multiplier for E.F. Empirical formula weight
The real usefulness of balanced equations can be Observed in the following series of predictions: I Predict moles of any products of a reaction II Predict the amount of any and all reactants needed for a given amount of product III Determine precisely how much of all reactants are used, and which are present in excess. The basis of chemical calculations for reactions Is the balanced reaction equation.
So far, we have seen mole ratio s from an equation That tells us how much to expect in the reaction: N 2 (g) + H 2 (g) NH 3 (g) N 2 (g) + 3H 2 (g) 2NH 3 (g) (Balanced equation) Think: mols reactant x moles product = mols product mols reactant What we are asked for goes on top of the ratio. What we Are given goes on bottom. How many moles of NH 3 can be produced from 5.00 mol H 2 5.00 mol H 2 (given) x 2 mol NH 3 = 3.33 mol NH 3 formed 3 mol H 2
In the laboratory, we must measure the mols by Mass. This requires an initial conversion to mols From the given mass of substance: How many moles of NH 3 are formed from 33.6 g Of N 2 : 33.6 g x 1 mol N 2 x 2 mol NH 3 = 2.40 mol NH 3 28.0 g N 2 1 mol N 2 Mass reactant x 1 x mol product = mol product F.W react. mol reactant
Finally, we must determine the mass of a product In the laboratory, since we cannot evaluate moles: What mass of H 2 is needed to produce 119 g of NH 3 What is asked for: mass of H 2 What is given: mass of NH 3 Mass reactant x 1 x mol product x FW prod =mass F.W react. mol reactant product 119 g NH 3 x1 mol NH 3 x 3 mol H 2 x 2.02 g H 2 = 21.2 g H 2 17.0 g NH 3 2 mol NH 3 mol H 2
Memorize these steps: Mass react x 1 mol react x mol product x prod. g = mass grams react mol react mol prod.
Mass react x 1 mol react x mol product x prod. g = mass grams react mol react mol prod. SiO 2 is etched with HF (aq): SiO 2 (s) + 4 HF (aq) SiF 4 (g) + 2 H 2 O (l) If there is 4.86 moles HF and 60 g SiO 2, will there be Any glass left over? 4.86 mol HF x 1 mol SiF 4 x 104 g SiF 4 = 126 g SiF 4 ; 4 mol HF mol SiF 4 126 g SiF 4 x 1 mol SiF 4 x SiO 2 x 60g SiO 2 = 72.7 g SiO 2 104 g SiF 4 SiF 4 mol SiO 2 Since I have only 60g (1 mol) of SiO 2, I will run out of Glass before I use up the acid! SiO 2 is LIMITING
4 Ag (s) + 2 H 2 S (g) + O 2 (g) 2 Ag 2 S (s) + 2 H 2 O (l) IF there is 0.145 mol Ag, 0.0872 mol H 2 S, and excess O 2, how much Ag 2 S is produced and what mass of Reactant is in excess? Ag: 0.145 mol x 2 mol Ag 2 S = 0.0725 mol Ag 2 S 4 mol Ag H 2 S: 0.0872 mol H 2 S x 2 mol Ag 2 S = 0.0872 mol Ag 2 S 2 mol H 2 S Ag is limiting and H 2 S is present in excess. We must use the limiting reagent quantity to predict actual products: 0.0725 mol Ag 2 S x 248 g Ag 2 S = 18.0 g Ag 2 S formed mol Ag 2 S
A reaction in the laboratory results in a product mass That is not quite what the balanced equation calculation Predicts. The difference between the theoretical amount and the Actual one is called Percent Yield: Actual Yield (mass) x 100% = percent yield Theoretical yield (mass)
2S (s) + 3O 2 (g) ->2 SO 3 (g) S O Reactant Mixture Product Mixture
Mass Changes in Chemical Reactions 1. Write balanced chemical equation 2. Convert quantities of known substances into moles 3. Use coefficients in balanced equation to calculate the number of moles of the sought quantity 4. Convert moles of sought quantity into desired units
Memorize these steps: mol ratio from balanced equation molar mass of product Mass react x 1 mol react x mol product x product g = mass grams react mol react mol prod. molar mass of reactant
Stoichiometric Relationships Molarity of reactant a,b,c or d From balanced equation Molarity of product Density (g/ml) (mol/g) Volume (L) Volume -1 (L -1 ) (g/mol) 1/density (ml/g) Volume reactant Grams reactant Moles reactant * moles product moles reactant * Moles product Grams of product Volume of product 1/N a N a Atoms or molecules Of reactant Atoms or molecules Of products a Reactant A + b Reactant B c Product C + d Product D or a Product A + b Product B c Reactant C + d Reactant D 2004 Randal Hallford
Methanol burns in air according to the equation 2CH 3 OH + 3O 2 2CO 2 + 4H 2 O If 209 g of methanol are used up in the combustion, what mass of water is produced? grams CH 3 OH moles CH 3 OH moles H 2 O grams H 2 O molar mass CH 3 OH coefficients chemical equation molar mass H 2 O 209 g CH 3 OH 1 mol CH 3 OH x 32.0 g CH 3 OH x 4 mol H 2 O 2 mol CH 3 OH 18.0 g H 2 O x = 1 mol H 2 O 235 g H 2 O
2S (s) + 3O 2 (g) ->2 SO 3 (g) Limiting Reagent S O Reactant Mixture Excess Reagent Product Mixture
Limiting Reagent Problems 1. Balance chemical equation 2. Determine limiting reagent Do two separate calculations for the amount of product each reactant would produce if they were the limiting reagent The reactant that gives the lower number is the limiting reagent. 3. The amount of product produced is the number calculated by limiting reagent
Limiting Reagent Problems 4. Determination of the amount of excess reagent left over Calculate the amount of excess reagent (ER) used in chemical reaction Subtract the ER used from original amount of ER.
Calculating Amounts of Reactant and Product in Reactions Involving a Limiting Reactant PROBLEM: A fuel mixture used in the early days of rocketry is composed of two liquids, hydrazine(n 2 H 4 ) and dinitrogen tetraoxide(n 2 O 4 ), which ignite on contact to form nitrogen gas and water vapor. How many grams of nitrogen gas form when 1.00x10 2 g of N 2 H 4 and 2.00x10 2 g of N 2 O 4 are mixed? PLAN: We always start with a balanced chemical equation and find the number of mols of reactants and products which have been given. In this case one of the reactants is in molar excess and the other will limit the extent of the reaction. mass of N 2 H 4 divide by M mass of N 2 O 4 limiting mol N 2 multiply by M mol of N 2 H 4 molar ratio mol of N 2 O 4 g N 2 mol of N 2 mol of N 2
Calculating Amounts of Reactant and Product in Reactions Involving a Limiting Reactant SOLUTION: 2 N 2 H 4 (l) + N 2 O 4 (l) 3 N 2 (g) + 4 H 2 O(l) 1.00x10 2 g N 2 H mol N 2 H 4 4 = 3.12mol N2 H 4 32.05g N 2 H 4 3.12mol N 2 H 4 3 mol N 2 = 4.68mol N 2 2mol N 2 H 4 mol 2.00x10 2 g N 2 O 4 N 2 O 4 = 2.17mol N 2 O 4 92.02g N 2 O 4 N 2 H 4 is the limiting reactant because it produces less product, N 2, than does N 2 O 4. 4.68mol N 2 mol N 2 28.02g N 2 = 131g N 2 2.17mol N 2 O 4 3 mol N 2 = 6.51mol N 2 mol N 2 O 4
Limiting Reagents 124 g of Al are reacts with 601 g of Fe 2 O 3 : 2Al + Fe 2 O 3 Al 2 O 3 + 2Fe Calculate the mass of Al 2 O 3 formed. g Al mol Al mol Fe 2 O 3 needed g Fe 2 O 3 needed OR g Fe 2 O 3 mol Fe 2 O 3 mol Al needed g Al needed 124 g Al x 1 mol Al 27.0 g Al x 1 mol Fe 2 O 3 2 mol Al 160. g Fe 2 O x 3 = 367 g Fe 1 mol Fe 2 O 2 O 3 3 Start with 124 g Al need 367 g Fe 2 O 3 Have more Fe 2 O 3 (601 g); Al is limiting reagent
Use limiting reagent (Al) to calculate amount of product that can be formed. g Al mol Al mol Al 2 O 3 g Al 2 O 3 2Al + Fe 2 O 3 Al 2 O 3 + 2Fe 124 g Al x 1 mol Al 27.0 g Al x 1 mol Al 2 O 3 2 mol Al 102. g Al 2 O x 3 = 234 g Al 1 mol Al 2 O 2 O 3 3
Theoretical Yield is the amount of product that would result if all the limiting reagent reacted. Actual Yield is the amount of product actually obtained from a reaction. % Yield = Actual Yield Theoretical Yield x 100
Calculating Percent Yield PROBLEM: PLAN: Silicon carbide (SiC) is an important ceramic material that is made by allowing sand(silicon dioxide, SiO2) to react with powdered carbon at high temperature. Carbon monoxide is also formed. When 100.0kg of sand are processed, 51.4kg of SiC are recovered. What is the percent yield of SiC in this process? SOLUTION: write balanced equation SiO 2 (s) + 3C(s) SiC(s) + 2CO(g) find mol reactant & product 100.0kg SiO 2 10 3 g SiO 2 kg SiO 2 mol SiO 2 = 1664 mol SiO 2 60.09g SiO 2 find g product predicted actual yield/theoretical yield x 100 percent yield mol SiO 2 = mol SiC = 1664 40.10g SiC kg 1664mol SiC mol SiC 10 3 g = 66.73kg 51.4kg x100 =77.0% 66.73kg
If a sample of acetylene is burned, and releases 550 KJ of Energy, how many grams of CO 2 can we expect to form? 2 C 2 H 2 (g) + 5O 2 (g) 4CO 2 (g) + 2H 2 O (l) H=-2602 KJ We can relate the moles of CO 2 to the amount of heat, since It is part of the balanced equation: 550 KJ x 4 mol CO 2 x 44.0 g CO 2 = 37.2 g CO 2 2602 KJ mol CO 2
Chemical Equations reactants products 3 ways of representing the reaction of H 2 with O 2 to form H 2 O
Balancing Chemical Equations 1. Write the correct formula(s) for the reactants on the left side and the correct formula(s) for the product(s) on the right side of the equation. Ethane reacts with oxygen to form carbon dioxide and water C 2 H 6 + O 2 CO 2 + H 2 O 2. Change the numbers in front of the formulas (coefficients) to make the number of atoms of each element the same on both sides of the equation. Do not change the subscripts. 2C 2 H 6 NOT C 4 H 12
Balancing Chemical Equations 3. Start by balancing those elements that appear in only one reactant and one product. C 2 H 6 + O 2 CO 2 + H 2 O start with C or H but not O 2 carbon on left C 2 H 6 + O 2 1 carbon on right 2CO 2 + H 2 O multiply CO 2 by 2 6 hydrogen on left C 2 H 6 + O 2 2 hydrogen on right 2CO 2 + 3H 2 O multiply H 2 O by 3
Balancing Chemical Equations 4. Balance those elements that appear in two or more reactants or products. C 2 H 6 + O 2 2CO 2 + 3H 2 O multiply O 2 by 7 2 2 oxygen on left 4 oxygen (2x2) + 3 oxygen (3x1) = 7 oxygen on right C 2 H 6 + 7 O 2 2CO 2 + 3H 2 O 2 remove fraction multiply both sides by 2 2C 2 H 6 + 7O 2 4CO 2 + 6H 2 O
Balancing Chemical Equations 5. Check to make sure that you have the same number of each type of atom on both sides of the equation. 2C 2 H 6 + 7O 2 4CO 2 + 6H 2 O 4 C (2 x 2) 4 C 12 H (2 x 6) 12 H (6 x 2) 14 O (7 x 2) 14 O (4 x 2 + 6) Reactants 4 C 12 H 14 O Products 4 C 12 H 14 O
I. Formulas show chemistry at a standstill. Equations show chemistry in action. A. Equations show: 1. the reactants which enter into a reaction. 2. the products which are formed by the reaction. 3. the amounts of each substance used and each substance produced. B. Two important principles to remember: 1. Every chemical compound has a formula which cannot be altered. 2. A chemical reaction must account for every atom that is used. This is an application of the Law of Conservation of Matter which states that in a chemical reaction atoms are neither created nor destroyed. C. Some things to remember about writing equations: 1. The diatomic elements when they stand alone are always written H 2, N 2, O 2, F 2, Cl 2, Br 2, I 2 2. The sign, ----->, means "yields" and shows the direction of the action. 3. A small delta, ( ), above the arrow shows that heat has been added. 4. A double arrow, <----->, shows that the reaction is reversible and can go in both directions. 5. Before beginning to balance an equation, check each formula to see that it is correct. NEVER change a formula during the balancing of an equation. 6. Balancing is done by placing coefficients in front of the formulas to insure the same number of atoms of each element on both sides of the arrow.
7. Always consult the Activity Series of metals and non-metals before attempting to write equations for replacement reactions. 8. If a reactant or product is solid, place (s) after the formula 9. If the reactant or product is a liquid, place (l) after the formula 10. If the reactant or product is a gas, place (g) after the formula 11. If the reactant or product is in water, place (aq) after the formula 12. A category of reaction will produce an unstable product which decomposes: H 2 CO 3 (aq) H 2 O (l) + CO 2 (g) carbonic acid H 2 SO 3 (aq) H 2 O (l) + SO 2 (g) sulfurous acid NH 4 OH (aq) NH 3 (g) + H 2 O (l) ammonium hydroxide
Rules for Writing Chemical Equations: 1. Write down the formula(s) for the reactants and add a + between them then put a yields arrow ( ) at the end 2. Examine the formula(s) to determine which of four basic types of reactions will occur. On the basis of your decision, write down correct formula(s) for the products on the right side of the yield arrow. 3. Place a coefficient in front of each species in the reactants and the products to ensure that the conservation of matter is observed: Balance the reaction
6. A few nonmetals combine with each other. 2P + 3Cl ----> 2PCl II. Four Basic Types of Chemical reactions A. Synthesis two or more elements or compounds combine to give a more complex product Examples of Synthesis reactions: 1. Metal + oxygen -----> metal oxide 2Mg (s) + O 2(g) ----> 2MgO (s) 2. Nonmetal + oxygen -----> nonmetallic oxide C (s) + O 2(g) ----> CO 2(g) 3. Metal oxide + water -----> metallic hydroxide MgO (s) + H 2 O (l) ----> Mg(OH) 2(s) 4. Nonmetallic oxide + water -----> acid CO 2(g) + H 2 O (l) ----> ; H 2 CO 3(aq) 5. Metal + nonmetal -----> salt 2 Na (s) + Cl 2(g) ----> 2NaCl (s)
B. Decomposition: A single compound breaks down into simpler compounds. Basic form: AX -----> A + X Examples of decomposition reactions (when heated): 1. Metallic carbonates form metallic oxides and CO 2(g). CaCO 3(s) ----> CaO (s) + CO 2(g) 2. Most metallic hydroxides decompose into metal oxides and water. Ca(OH) 2(s) ----> CaO (s) + H 2 O (g) 3. Metallic chlorates decompose into metallic chlorides and oxygen. 2KClO 3(s) ----> 2KCl (s) + 3O 2(g) 4. Some acids decompose into nonmetallic oxides and water. H 2 SO 4 ----> H 2 O (l) + SO 3(g) 5. Some oxides decompose. 2HgO (s) ----> 2Hg (l) + O 2(g) 6. Some decomposition reactions are produced by electricity. 2H 2 O (l) ----> 2H 2(g) + O 2(g) 2NaCl (l) ----> 2Na (s) + Cl 2(g)
C. Replacement: a more active element takes the place of another element and frees the less active one. Basic form: A + BX -----> AX + B or AX + Y -----> AY + X Examples of replacement reactions: 1. Replacement of a metal in a compound by a more active metal. Fe (s) + CuSO 4(aq) ----> FeSO 4(aq) + Cu (s) 2. Replacement of hydrogen in water by an active metal. 2Na (s) + 2H 2 O (l) ----> 2NaOH (aq) + H 2(g) Mg (s) + H 2 O (g) ----> MgO (s) + H 2(g) 3. Replacement of hydrogen in acids by active metals. Zn (s) + 2HCl (aq) ----> ZnCl 2(aq) + H 2(g) 4. Replacement of nonmetals by more active nonmetals. Cl 2(g) + 2NaBr (aq) ----> 2NaCl (aq) + Br 2(l) NOTE: Refer to the Activity Series for metals and nonmetals to predict products of replacement reactions. If the free element is above the element to be replaced in the compound, then the reaction will occur. If it is below, then no reaction occurs.
If the free element is above the element to be replaced in the compound, Then the reaction will occur. If the free element is below it, no reaction occurs.
D. Ionic or Double Displacement: occurs between ions in aqueous solution. A reaction will occur when a pair of ions come together to produce at least one of the following: 1. a precipitate 2. a gas 3. water or some other non-ionized substance. Basic form: AX + BY -----> AY + BX Examples of ionic reactions: 1. Formation of precipitate. NaCl (aq) + AgNO 3(aq) ----> NaNO 3(aq) + AgCl (s) BaCl 2(aq) + Na 2 SO 4(aq) ----> 2NaCl (aq) + BaSO 4(s) 2. Formation of a gas. HCl (aq) + FeS (s) ----> FeCl 2(aq) + H 2 S (g) 3. Formation of water. (If the reaction is between an acid and a base it is called a neutralization reaction.) HCl (aq) + NaOH (aq) ----> NaCl (aq) + H 2 O (l) 4. Formation of a product which decomposes. CaCO 3(s) + HCl (aq) ----> CaCl 2(aq) + CO 2(g) + H 2 O (l)
Use the Solubility Rules to decide whether a product of an ionic reaction is insoluble in water and will thus form a precipitate. If a compound is soluble in water then it should be shown as being in aqueous solution, or left as separate ions. It is, in fact, often more desirable to show only those ions that are actually taking part in the actual reaction. Equations of this type are called Net Ionic equations.
Combustion of Hydrocarbons: Another important type of reaction, in addition to the four types above, is that of the combustion of a hydrocarbon. When a hydrocarbon is burned with sufficient oxygen supply, the products are always carbon dioxide and water vapor. If the supply of oxygen is low or restricted, then carbon monoxide will be produced. This is why it is so dangerous to have an automobile engine running inside a closed garage or to use a charcoal grill indoors. Hydrocarbon (C x H y ) + O 2(g) -----> CO 2(g) + H 2 O (g) CH 4(g) + 2O 2(g) ----> CO 2(g) + 2H 2 O (g) 2C 4 H 10(g) + 13O 2(g) ----> 8CO 2(g) + 10H 2 O (g) NOTE: Complete combustion means the higher oxidation number is attained. Incomplete combustion means the lower oxidation number is attained. The phrase "To burn" means to add oxygen unless told otherwise.
Conservation of matter requires identical numbers of atoms on both sides Begin by inspecting the number of each type of atom on each side. Nitrogen must separate before the reaction Hydrogen must separate before the reaction. Re-combine in a different chemical species Same atoms, different ratio!
D. Ionic or Double Displacement: occurs between ions in aqueous solution. A reaction will occur when a pair of ions come together to produce at least one of the following: 1. a precipitate 2. a gas 3. water or some other non-ionized substance. Basic form: AX + BY -----> AY + BX Examples of ionic reactions: 1. Formation of precipitate. NaCl (aq) + AgNO 3(aq) ----> NaNO 3(aq) + AgCl (s) BaCl 2(aq) + Na 2 SO 4(aq) ----> 2NaCl (aq) + BaSO 4(s) 2. Formation of a gas. HCl (aq) + FeS (s) ----> FeCl 2(aq) + H 2 S (g) 3. Formation of water. (If the reaction is between an acid and a base it is called a neutralization reaction.) HCl (aq) + NaOH (aq) ----> NaCl (aq) + H 2 O (l) 4. Formation of a product which decomposes. CaCO 3(s) + HCl (aq) ----> CaCl 2(aq) + CO 2(g) + H 2 O (l)
Use the Solubility Rules to decide whether a product of an ionic reaction is insoluble in water and will thus form a precipitate. If a compound is soluble in water then it should be shown as being in aqueous solution, or left as separate ions. It is, in fact, often more desirable to show only those ions that are actually taking part in the actual reaction. Equations of this type are called Net Ionic equations.