Chapter 3 Molecules, Compounds, and Chemical Equations

3.7 Formula Mass versus Molar mass The average mass of a molecule or formula unit in amu also known as molecular mass or molecular weight (MW) whole = sum of the parts! Molar mass Total mass of a compound in gram per 1 mol of its molecules or formula unit

Molar Mass of Compounds the relative masses of molecules can be calculated from atomic masses Formula Mass = 1 molecule of H2O = 2(1.01 amu H) + 16.00 amu O = 18.02 amu since 1 mole of H2O contains 2 moles of H and 1 mole of O Molar Mass = 1 mole H2O = 2(1.01 g H) + 16.00 g O = 18.02 g so the Molar Mass of H2O is 18.02 g/mole Why multiplying 2?

Molar Mass of Na2SO4 Calculate the molar mass of Na2SO4. Element Number of Moles Atomic Mass Total Mass in each element Na S O Total mass in 1 mol Na2SO4

3.8 Mass Percent Composition Percentage of each element in a compound By mass Can be determined from the formula of the compound the experimental mass analysis of the compound The percentages may not always total to 100% due to rounding

Example - Mass Percent as a Conversion Factor Calculate the mass percent of Na in NaCl Benzaldehyde is 79.2% carbon. What mass of benzaldehyde contains 19.8 g of C?

Chemical Formulas and Elemental Composition chemical formulas have inherent in them relationships between numbers of atoms and molecules or moles of atoms and molecules these relationships can be used to convert between amounts of constituent elements and molecules like percent composition Vol A  Grams A  Moles A  Moles B  Grams B Grams A  Moles A  Moles B  Grams B

Example Butane (C4H10) is the liquid fuel in lighters. a. Determine the number of atoms ratio between carbon and 1 molecule of C4H10 Determine the number moles ratio between C and 1 mol of C4H10 How many grams of carbon are present within a lighter containing 7.5 mL of butane? The density of quid butane is 0.601 g/mL

Empirical Formula simplest, whole-number ratio of the atoms of elements in a compound can be determined from elemental analysis masses of elements formed when decompose or react compound combustion analysis percent composition

Steps in determine the Empirical formula Step 1: Obtain the mass of each element (in grams) E.G 100% = 100g therefore mass percent is the same numerical value in grams Step 2: Determine the numbers moles of each atom present Use molar mass of each element Step 3: Divide the smallest moles by numbers moles of each atom to obtain the closet integer as possible. if result is within 0.1 of whole number, round to whole number Step 4: If the result ended with 0.5, 0.33, 1.125, 1.50 etc… then multiply with a factor to get the nearest integer as possible. E.g 1.5 x 2 = 3.0 atoms 1.33 x 3 = 3. 99 = 4 atoms Step 5: Write the result (number atoms) from step 4 as a subscript for the appropriate element.

Example Determine the empirical formula of stannous fluoride, which contains 75.7% Sn (118.70g/mol) and the rest fluorine (19.00 g/mol) An unknown sample gives the following mass percent: 17.5% Na, 39.7% Cr and 42.8% O. What is the empirical formula?

Molecular Formulas The molecular formula is a multiple of the empirical formula To determine the molecular formula you need to know the empirical formula and the molar mass of the compound molecular mass empirical formula mass Multiple (n) = Molecular formula = empirical formula x n where n = 1, 2, 3, 4

Example Laboratory analysis of aspirin determined the following mass percent composition. Find the empirical formula and molecular formula C = 60.00% H = 4.48% O = 35.53%

Determining Empirical Formulas: Elemental Analysis Chapter 3: Formulas, Equations, and Moles Determining Empirical Formulas: Elemental Analysis 11/22/2018 Combustion Analysis: A compound of unknown composition (containing a combination of carbon, hydrogen, and possibly oxygen) is burned with oxygen to produce the volatile combustion products CO2 and H2O, which are separated and weighed by an automated instrument called a gas chromatograph. hydrocarbon + O2(g) xCO2(g) + yH2O(g) carbon hydrogen If the unknown compound also contains oxygen, then the masses of carbon and hydrogen are determined first. The mass of oxygen is determined by subtracting those masses from that of the unknown compound. Copyright © 2008 Pearson Prentice Hall, Inc.

Combustion Analysis Unknown formula: CxHyOx (Oxygen can be replaced with other nonmetal) gCO2  moles CO2  moles C  gC gH2O  moles H2O  moles H  gH g O = g sample – (g H + g C) gO  moles O Follow steps in determine the empirical formula and molecular formula

Example Combustion of a 0.8233 g sample of a compound containing only carbon, hydrogen, and oxygen produced the following: CO2 = 2.445 g H2O = 0.6003 g Determine the empirical formula of the compound

Example Upon combustion, a compound containing only carbon and hydrogen produced 1.60g CO2 and 0.819g H2O. Find the empirical formula

Chemical Reactions Reactions involve chemical changes in matter resulting in new substances Reactions involve rearrangement and exchange of atoms to produce new molecules Elements are not transmuted during a reaction

Chemical Equations A chemical equation gives the formulas of the reactants on the left of the arrow. the formulas of the products on the right of the arrow. Reactants Product O2 (g) CO2 (g) C(s)

Symbols Used in Equations Symbols in chemical equations show the states of the reactants. the states of the products. the reaction conditions. TABLE

Chemical Equations are Balanced In a balanced chemical reaction no atoms are lost or gained. the number of reacting atoms is equal to the number of product atoms.

Balancing Chemical Equations Chapter 3: Formulas, Equations, and Moles 11/22/2018 Balancing Chemical Equations A balanced chemical equation shows that the law of conservation of mass is adhered to. In a balanced chemical equation, the numbers and kinds of atoms on both sides of the reaction arrow are identical. 2NaCl(s) 2Na(s) + Cl2(g) left side: 2 Na 2 Cl right side: 2 Na 2 Cl Remember, a chemical reaction can be thought of as a rearrangement of the atoms to form new compounds. Copyright © 2008 Pearson Prentice Hall, Inc.

Balancing Chemical Equations Chapter 3: Formulas, Equations, and Moles Balancing Chemical Equations 11/22/2018 Write the unbalanced equation using the correct chemical formula for each reactant and product. H2O(l) H2(g) + O2(g) Find suitable coefficients—the numbers placed before formulas to indicate how many formula units of each substance are required to balance the equation. 2H2O(l) 2H2(g) + O2(g) Reduce the coefficients to their smallest whole-number values, if necessary, by dividing them all by a common denominator. 2H2O(l) 2H2(g) + O2(g) Copyright © 2008 Pearson Prentice Hall, Inc.

Balancing Chemical Equations Chapter 3: Formulas, Equations, and Moles 11/22/2018 Balancing Chemical Equations Check your answer by making sure that the numbers and kinds of atoms are the same on both sides of the equation. 2H2O(l) 2H2(g) + O2(g) left side: 4 H 2 O right side: 4 H 2 O Copyright © 2008 Pearson Prentice Hall, Inc.

Balancing Chemical Equations Chapter 3: Formulas, Equations, and Moles Balancing Chemical Equations 11/22/2018 Do not change subscripts when you balance a chemical equation. You are only allowed to change the coefficients. H2O(l) H2(g) + O2(g) unbalanced 2H2O(l) 2H2(g) + O2(g) H2O2(l) H2(g) + O2(g) Balancing chemical equations only determines the proper numerical ratios between all of the substances (both reactants and products). It does not change any of the substances. Balanced properly Chemical equation changed! Copyright © 2008 Pearson Prentice Hall, Inc.

Examples Balance the coefficients from reactants to products. __N2(g) + __H2(g) __ NH3(g) B. __Co2O3(s) + __ C(s) __Co(s) + __CO2(g) Write a balanced equation for the reaction between carbon dioxide gas and aqueous potassium hydroxide to form potassium carbonate and water. The combustion of gaseous ethane (C2H6)