25 5.2 Molar Mass

Practice questions

Multiple choice questions

1. How many acetaminophen molecules are present in 0.25mol of acetaminophen?
   1. 24.088 × 10²³ acetaminophen molecules
   2. 1.5 × 10²³ acetaminophen molecules
   3. 0.042 × 10²³ acetaminophen molecules
   4. 6.022 × 10²³ acetaminophen molecules
2. Glucose has the formula C₆H₁₂O₆. Calculate the molar mass of glucose. The molar masses of C, H, and O are given here: C = 12.01g/mol, H = 1.01g/mol, O = 16.00g/mol
   1. 174.12g/mol
   2. 29.02g/mol
   3. 53.02g/mol
   4. 180.1g/mol
3. How many moles of C are present in 1.5mol of glucose? The chemical formula of glucose is C₆H₁₂O₆.
   1. 9mol
   2. 18mol
   3. 4mol
   4. 0.25mol
4. How many moles of H are present in 1.5mol of glucose? The chemical formula of glucose is C₆H₁₂O₆.
   1. 9mol
   2. 18mol
   3. 4mol
   4. 0.25mol
5. How many moles of O are present in 1.5mol of glucose? The chemical formula of glucose is C₆H₁₂O₆.
   1. 9mol
   2. 18mol
   3. 4mol
   4. 0.25mol

Short-answer questions

1. Calculate the amount of molecular oxygen O₂ present in 160.0g of oxygen. The molar mass of atomic O is 16.00g/mol.
   Step 1: Calculate the molar mass of molecular oxygen: ________g/mol× _________=_____________ g/mol
   Step 2: Calculate the moles of molecular oxygen (HINT: Molar Mass= Mass ÷ Moles): Moles n =_____________________÷________________= ________________ mol
2. Calculate how many grams of carbon are present in 1.8g of glucose.
   The chemical formula of glucose is C₆H₁₂O₆. The atomic molar masses of C, H and O are given:
   C = 12.01g/mol, H = 1.01g/mol, O = 16.00g/mol. Please round the answers to two d.p.
   1. Step 1 – Calculate the molar mass of glucose 12.01g/mol× _______+1.01g/mol× _________ +16.00g/mol×________ = ________ g/mol
   2. Step 2 – Calculate the number of moles of glucose present in 1.8g of glucose n =m/M
      n = _______ ÷180.1 = __________ mol </li?
   3. Step 3 – Calculate the number of moles of carbon present in the number of moles of glucose from step 2. 6mol C in 1mol C₆H₁₂O₆ × _________ mol C₆H₁₂O₆ = _________ mol  C
   4. Step 4 – Convert the number of moles of carbon into grams m = n × M
      m = _________ ×12.01g/mol
      m = ________ g

Solutions

Multiple choice questions

1. b
2. d Glucose contains six atoms of C, twelve atoms of H and six atoms of O. (6 × 12.01g/mol) + (12 × 1.01g/mol) + (6 × 16.00g/mol) = 180.1g/mol.
3. a
4. b
5. a

Short-answer questions

1. Step 1: 16g/mol×2 = 32g/mol
   Step 2: Moles n = 160 ÷ 32 = 5mol
2. Step 1: 12.01g/mol×6+1.01g/mol×12+16.00g/mol×6=180.10g/mol
   Step 2: n = 1.8÷180.1 =0.01mol
   Step 3: 6mol C in 1mol C₆H₁₂O₆× 0.01mol C₆H₁₂O₆ =0.06mol C
   Step 4: m = n × M = 0.06×12.01g/mol = 0.72 g