# 1. A sample of gold (Au) has a mass of 35.12 g. a. Calculate the number of moles of gold (Au) in the sample and record in Table 1. Show your work. b. Calculate the number of atoms of gold (Au) in the sample and record in Table 1. Show your work. 2. A sample of table sugar (sucrose, C12H22O11) has a mass of 1.202 g. a. Calculate the number of moles of C12H22O11 contained in the sample and record in Table 1. Show your work. b. Calculate the moles of each element in C12H22O11 and record in Table 1. Show your work. c. Calculate the number of atoms of each type in C12H22O11 and record in Table 1. Show your work.

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/1M A sample of gold: xM = 35.12g/1000ml ⇒ 1000 ml = 35.12g/x 1000 ml = 196.97g/1M = 35.12g/x ⇒ 196.97g/1M = 35.12g/x x = 35.12g / 196.97g * 1M = 0.1783M Q1. 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 M After crossing the products:x = 6.02× 10²³atoms * 0.1783 M / 1M = 1.073 × 10²³ atoms Q2. 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 M Q2. 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 M Q2. 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 M After 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 M After 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 M After crossing the products:x = 6.02× 10²³atoms * 0.0385 M / 1M = 0.23 × 10²³ atoms = 2.3 × 10²⁴ atoms