Q1. a sample of gold (au) has a mass of 45.65 g. • a) calculate the number of moles of gold (au) in the sample. show your work and highlight your final answer. b) calculate the number of atoms of gold (au) in the sample. (use your answer from part a.) show your work and highlight your final answer. q2. a sample of sodium chloride (nacl) has a mass of 28.36 g. • a) calculate the number of moles of nacl in the sample. show your work and highlight your final answer. b) calculate the number of moles of each element (number of moles of na & number of moles of cl) in nacl in the sample. (use your answer from part a.) show your work and highlight your final answer. c) calculate the number of atoms of each element (number of atoms of na & number of atoms of cl) in nacl in the sample. (use your answers from part b.) show your work and highlight your final answer.

Q1. a) the answer is 0.1783 moles.atomic mass (ar) of gold is 196.97 g. a mole (m) is an atomic or molar mass in 1000 ml: au: 1m = 196.97g/1000ml                            ⇒ 1000 ml = 196.97g/1ma sample of gold: xm = 35.12g/1000ml      ⇒ 1000 ml = 35.12g/x1000 ml = 196.97g/1m = 35.12g/x⇒ 196.97g/1m = 35.12g/x      x = 35.12g / 196.97g * 1m = 0.1783mq1. b) the answer is 1.073 × 10²³ atoms.to calculate this, we will use avogadro's number which is the number of units (atoms, molecules) in 1 mole of substance: 6.02 × 10²³ units per 1 molefrom the previous task, we know that the sample of gold has 0.1783 moles.now, let's make a proportion: 6.02 × 10²³atoms : 1m = x : 0.1783 mafter crossing the products: x = 6.02× 10²³atoms * 0.1783 m / 1m = 1.073 × 10²³ atomsq2. a) the answer is 0.0035 moles.let's first calculate molar mass (mr) of sucrose which is a sum of atomic masses (ar) of elements: mr(c₁₂h₂₂o₁₁) = 12ar(c) + 22ar(h) + 11ar(o) = 12*12 + 22*1 + 11*16 =                        = 144 + 22 + 176 = 342 ga mole (m) is an atomic or molar mass in 1000 ml: sucrose: 1m = 342g/1000ml                                ⇒ 1000 ml = 342g/1ma sample of sucrose: xm = 1.202g/1000ml        ⇒ 1000 ml = 1.202g/x1000 ml = 342g/1m = 1.202g/x⇒  342g/1m = 1.202g/x      x = 1.202g / 342g * 1m = 0.0035 mq2. b) the answers are: - carbon: 0.042 moles- hydrogen: 0.077 moles- oxygen: 0.0385 molesin a sample of sucrose of 0.0035 m, there are 12 atoms of carbon: 12 * 0.0035m = 0.042 min a sample of sucrose of 0.0035 m, there are 22 atoms of hydrogen: 22 * 0.0035m = 0.077 min a sample of sucrose of 0.0035 m, there are 11 atoms of oxygen: 11 * 0.0035m = 0.0385 mq2. c) the answers are: - carbon: 2.5 × 10²⁴ atoms- hydrogen: 4.6 × 10²⁴ atoms- oxygen: 2.3 × 10²⁴ atomsto calculate this, we will use avogadro's number which is the number of units (atoms, molecules) in 1 mole of substance: 6.02 × 10²³ units per 1 mole- carbon: 0.042 moles (from the previous task)now, let's make a proportion: 6.02 × 10²³atoms : 1m = x : 0.042 mafter crossing the products: x = 6.02× 10²³atoms * 0.042 m / 1m = 0.25 × 10²³ atoms = 2.5 × 10²⁴ atoms- hydrogen: 0.077 moles (from the previous task)now, let's make a proportion: 6.02 × 10²³atoms : 1m = x : 0.077 mafter crossing the products: x = 6.02× 10²³atoms * 0.077 m / 1m = 0.46 × 10²³ atoms = 4.6 × 10²⁴ atoms- oxygen: 0.0385 moles (from the previous task)now, let's make a proportion: 6.02 × 10²³atoms : 1m = x : 0.0385 mafter crossing the products: x = 6.02× 10²³atoms * 0.0385 m / 1m = 0.23 × 10²³ atoms = 2.3 × 10²⁴ atoms  read more on -

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