Problem 3.14 How many moles of cobalt (Co) atoms are there in 6.00 X109 (6 billion) Co atoms?

PROBLEM 3.24-F CALCULATE THE MOLAR MASS OF: (f) Mg3N2

PROBLEM 3.27 CALCULATE THE NUMBER OF C, H, AND O ATOMS IN 1.50 G OF OF GLUCOSE (C6H12O6)

PROBLEM 3.30 The density of water is 1.00 g/ml at 4 Celsius degrees. How many water molecules are in 2.56 ml of water at this temperature?

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 x (12.01 g) 46.07 g x 100% = 52.14% C2H6O %H = 6 x (1.008 g) 46.07 g x 100% = 13.13% %O = 1 x (16.00 g) 46.07 g x 100% = 34.73% 52.14% + 13.13% + 34.73% = 100.0%

MASS RELATIONSHIPS: PERCENT COMPOSITION AND FORMULAS What is the percent of Hydrogen in H2O?

TYPES OF FORMULAS MOLECULAR (C6H1206) EMPIRICAL (CH2O)

Problem 3.39 Tin (Sn) exists in Earth’s crust as SnO2. Calculate the percent composition by mass of Sn and O in SnO2.

Determine the empirical formula of a compound that has the following percent composition by mass: K 24.75, Mn 34.77, O 40.51 percent.

PROCEDURE Find grams of each element. CONVERT GRAMS TO MOLES DIVIDE BY SMALLEST NUMBER OF MOLES EXPRESS MOLE RATIO

Determine the empirical formula of a compound that has the following percent composition by mass: K 24.75, Mn 34.77, O 40.51 percent.

PROBLEM 3.49 A, B WHAT ARE THE EMPIRICAL FORMULAS OF THE COMPOUNDS WITH THE FOLLOWING COMPOSITIONS? (a) 2.1 PERCENT H, 65.3 PERCENT O, 32.6 PERCENT S; (b) 20.2 PERCENT Al, 79.8 PERCENT Cl

Percent Composition and Empirical Formulas Determine the empirical formula of a compound that has the following percent composition by mass: K 24.75, Mn 34.77, O 40.51 percent. nK = 24.75 g K x = 0.6330 mol K 1 mol K 39.10 g K nMn = 34.77 g Mn x = 0.6329 mol Mn 1 mol Mn 54.94 g Mn nO = 40.51 g O x = 2.532 mol O 1 mol O 16.00 g O

PROBLEM 3.49 WHAT IS THE EMPIRICAL FORMULA OF A COMPOUND CONTAINING (A) 2.1 Percent H; 65.3 Percent O; 32.6 percent S

3 ways of representing the reaction of H2 with O2 to form H2O A process in which one or more substances is changed into one or more new substances is a chemical reaction A chemical equation uses chemical symbols to show what happens during a chemical reaction reactants products 3 ways of representing the reaction of H2 with O2 to form H2O

Balancing Chemical Equations 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 C2H6 + O2 CO2 + H2O 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. 2C2H6 NOT C4H12

Balancing Chemical Equations Start by balancing those elements that appear in only one reactant and one product. C2H6 + O2 CO2 + H2O start with C or H but not O 1 carbon on right 2 carbon on left multiply CO2 by 2 C2H6 + O2 2CO2 + H2O 6 hydrogen on left 2 hydrogen on right multiply H2O by 3 C2H6 + O2 2CO2 + 3H2O

Balancing Chemical Equations Balance those elements that appear in two or more reactants or products. multiply O2 by 7 2 C2H6 + O2 2CO2 + 3H2O 2 oxygen on left 4 oxygen (2x2) + 3 oxygen (3x1) = 7 oxygen on right C2H6 + O2 2CO2 + 3H2O 7 2 remove fraction multiply both sides by 2 2C2H6 + 7O2 4CO2 + 6H2O

Balancing Chemical Equations Check to make sure that you have the same number of each type of atom on both sides of the equation. 2C2H6 + 7O2 4CO2 + 6H2O 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 Products 4 C 12 H 14 O

PROBLEM 3.59 BALANCE THE FOLLOWING EQUATIONS. C + O2  CO H2 + Br2  HBr K + H2O  KOH + H2 H2O2  H2O + O2 S8 + O2  SO2 KOH + H3PO4  K3PO4 + H2O CH4 + Br2  CBr4 + HBr

Problem 3.59 BALANCE THE FOLLOWING EQUATIONS C + O2  CO H2 + Br2  HBr Zn + AgCl  ZnCl2 +Ag (k) NaOH + H2SO4  Na2SO4 + H2O

How to “Read” Chemical Equations 2 Mg + O2 2 MgO 2 atoms Mg + 1 molecule O2 makes 2 formula units MgO 2 moles Mg + 1 mole O2 makes 2 moles MgO 48.6 grams Mg + 32.0 grams O2 makes 80.6 g MgO NOT 2 grams Mg + 1 gram O2 makes 2 g MgO

Amounts of Reactants and Products Chemical Equations: Amounts of Reactants and Products Write balanced chemical equation Convert quantities of known substances into moles Use coefficients in balanced equation to calculate the number of moles of the sought quantity Convert moles of sought quantity into desired units

Problem 3.66 Silicon tetrachloride (SiCl4) can be prepared by heating Si in chlorine gas: Si(s) + Cl2(g) SiCl4(l) In one reaction 0.507 mole of SiCl4 is produced. How many moles of molecular chlorine were used in the reaction.

Problem 3.68 Consider the combustion of butane (C4H10 ): 2C4H10 + 13(O2)  8CO2 + 10H2O In a particular reaction, 5.0 moles of C4H10 are reacted with an excess of O2. Calculate the moles of CO2 formed.

Problem 3.71 When Potassium Cyanide (KCN) reacts with acids, a deadly poisonous gas, hydrogen cyanide (HCN) is given off. Here is the equation. KCN(aq)+ HCl(aq)  KCl(aq) + HCN(g) If a sample of 0.140 grams of KCN is treated with an excess of HCl, calculate the amount of HCN formed in grams.

Problem 3.75 Limestone (CaCO3) is decomposed by heating to quicklime (CaO) and carbon dioxide. Calculate how many grams of quicklime can be produced from 1.0 kg of limestone. CaCO3  CaO + CO2

Limiting Reagent: Reactant used up first in the reaction. 2NO + O2 2NO2 NO is the limiting reagent O2 is the excess reagent

In one process, 124 g of Al are reacted with 601 g of Fe2O3. 2Al + Fe2O3  Al2O3 + 2Fe Calculate the mass of Al2O3 formed.

Have more Fe2O3 (601 g) so Al is limiting reagent In one process, 124 g of Al are reacted with 601 g of Fe2O3 2Al + Fe2O3 Al2O3 + 2Fe Calculate the mass of Al2O3 formed. g Al mol Al mol Fe2O3 needed g Fe2O3 needed OR g Fe2O3 mol Fe2O3 mol Al needed g Al needed 1 mol Al 27.0 g Al x 1 mol Fe2O3 2 mol Al x 160. g Fe2O3 1 mol Fe2O3 x = 124 g Al 367 g Fe2O3 Start with 124 g Al need 367 g Fe2O3 Have more Fe2O3 (601 g) so Al is limiting reagent

1 mol Al 27.0 g Al x 1 mol Al2O3 2 mol Al x 102. g Al2O3 1 mol Al2O3 x Use limiting reagent (Al) to calculate amount of product that can be formed. g Al mol Al mol Al2O3 g Al2O3 2Al + Fe2O3 Al2O3 + 2Fe 1 mol Al 27.0 g Al x 1 mol Al2O3 2 mol Al x 102. g Al2O3 1 mol Al2O3 x = 124 g Al 234 g Al2O3 At this point, all the Al is consumed and Fe2O3 remains in excess.

PROBLEM 3.83 Nitric Oxide (NO) reacts with oxygen gas to form nitrogen dioxide (NO2), a dark brown gas: 2NO(g) + O2(g)  2NO2(g) In one experiment, 0.886 moles of NO are mixed with 0.503 moles of O2(g). Calculate which of the two is the limiting reagent. Calculate the moles of NO2 produced.

Reaction Yield 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%

Percent Yield In an organic chemistry lab, 12.4 grams of an organic compound was produced. According to calculations, the reaction should have produced 17.0 grams. Calculate the percent yield.

Chemistry In Action: Chemical Fertilizers Plants need: N, P, K, Ca, S, & Mg 3H2 (g) + N2 (g) 2NH3 (g) NH3 (aq) + HNO3 (aq) NH4NO3 (aq) 2Ca5(PO4)3F (s) + 7H2SO4 (aq) 3Ca(H2PO4)2 (aq) + 7CaSO4 (aq) + 2HF (g) fluorapatite