Chem1001 Worksheet 3: 5 Key Answers Revealed

In the journey of mastering chemistry, particularly for those enrolled in Chem1001, understanding worksheets is a pivotal step towards success. This blog post aims to shed light on the 5 key answers from Worksheet 3, offering not just solutions, but also insights into the underlying chemistry principles that could help you in your future studies.

Understanding Molecular Mass

One of the fundamental aspects of chemistry is the calculation of molecular mass, which is crucial for various analytical and stoichiometric computations. Let’s delve into this concept with an example from Worksheet 3:

• Question: Calculate the molecular mass of glucose (C₆H₁₂O₆).
• Answer: Molecular mass of glucose is calculated as follows:
  • Carbon ©: 6 × 12.01 g/mol = 72.06 g/mol
  • Hydrogen (H): 12 × 1.008 g/mol = 12.10 g/mol
  • Oxygen (O): 6 × 16.00 g/mol = 96.00 g/mol
  • Total: 72.06 + 12.10 + 96.00 = 180.16 g/mol

⚗️ Note: Remember to use periodic table values for precision, ensuring accuracy in your calculations.

Stoichiometry Fundamentals

Stoichiometry, the calculation of reactants and products in chemical reactions, is another essential topic covered in Worksheet 3. Here’s an example:

• Question: How many moles of oxygen are needed to react completely with 2 moles of methane (CH₄) in the combustion reaction?
• Answer:
  • The balanced equation for the combustion of methane is: CH₄ + 2O₂ → CO₂ + 2H₂O
  • From the equation, we see that 1 mole of methane reacts with 2 moles of oxygen. Therefore, for 2 moles of methane, 4 moles of oxygen are needed.

Empirical and Molecular Formulas

Determining empirical and molecular formulas from mass data is a skill that’s tested frequently in chemistry. Here’s how you approach it:

• Question: Given the mass composition of a compound is 40% carbon, 6.67% hydrogen, and 53.33% oxygen, find the empirical formula.
• Answer:
  • Convert the mass percentages into moles:
    • Carbon: ⁴⁰⁄₁₂.01 = 3.33 mol
    • Hydrogen: 6.⁶⁷⁄₁.008 = 6.62 mol
    • Oxygen: 53.³³⁄₁₆.00 = 3.33 mol
  • Determine the simplest whole number ratio:
    • Divide by the smallest number (3.33):
    • Carbon: 3.³³⁄₃.33 = 1
    • Hydrogen: 6.⁶²⁄₃.33 = 2
    • Oxygen: 3.³³⁄₃.33 = 1
  • Empirical Formula: CH₂O

Limiting Reagent and Percent Yield

Understanding which reagent limits the reaction and calculating percent yield are vital for industrial chemistry and lab work:

• Question: You mix 25g of magnesium with 15g of oxygen. Which is the limiting reagent in the formation of magnesium oxide?
• Answer:
  • Magnesium (Mg): 25g / 24.31 g/mol = 1.028 mol
  • Oxygen (O₂): 15g / 32.00 g/mol = 0.469 mol
  • The balanced equation is: 2Mg + O₂ → 2MgO
  • From this, we need twice as many moles of Mg as O₂. Here, magnesium is in excess. Thus, oxygen is the limiting reagent.

Understanding Concentration and Dilution

Concentration calculations and understanding the effects of dilution are key for any chemistry student:

• Question: If you dilute 250mL of 6M NaOH solution to make 1L, what is the final molarity?
• Answer:
  • Use the dilution formula: C₁V₁ = C₂V₂
  • Initial concentration (C₁) = 6M, Initial volume (V₁) = 0.250L, Final volume (V₂) = 1L
  • C₂ = (6 * 0.250) / 1 = 1.5M

To conclude, Worksheet 3 in Chem1001 offers a wide array of foundational chemistry concepts ranging from basic stoichiometry to complex molecular weight calculations. By mastering these principles through consistent practice and understanding, students can equip themselves with the knowledge necessary for more advanced studies in chemistry. The examples provided not only serve as solutions but as stepping stones for deeper learning and application of chemical principles.

What’s the difference between empirical and molecular formulas?

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The empirical formula gives the simplest whole number ratio of atoms in a compound, whereas the molecular formula shows the actual number of atoms in a molecule, which might be a multiple of the empirical formula.

Why is understanding molecular mass important in chemistry?

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Molecular mass helps in calculating the amount of substance in moles, which is fundamental for stoichiometry, reaction prediction, and preparation of solutions with specific concentrations.

Can I predict the outcome of a reaction with stoichiometry alone?

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Yes, with accurate stoichiometric calculations, you can predict how much of each reactant is needed and how much product can be formed, assuming the reaction goes to completion.