Empirical and molecular calculations and formular
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Questions and Answers

What formula can be used to determine the number of empirical formula units per molecule?

  • Molecular mass / Empirical formula mass (correct)
  • Empirical formula mass * Molar mass
  • Molar mass * Empirical formula mass
  • Empirical formula mass / Molecular mass

    • If a covalent compound's empirical formula is CH2O and its molecular mass is 180 amu, how many formula units are in each molecule?

  • 4
  • 2
  • 6 (correct)
  • 8

    • Why is it necessary to multiply each subscript in the empirical formula by n to obtain the molecular formula?

  • To increase the molecular mass
  • To decrease the number of atoms in the molecule
  • To match the molar mass with the empirical formula mass
  • To represent the actual number of atoms in the molecule (correct)

    • What does a C5H7N empirical formula for nicotine indicate?

    <p>The compound consists of 5 carbon atoms, 7 hydrogen atoms, and 1 nitrogen atom.</p> Signup and view all the answers

    What is the molar ratio of hydrogen to nitrogen in the empirical formula C5H7N for nicotine?

    <p>7:1</p> Signup and view all the answers

    Why is comparison of molar mass to empirical formula mass important in determining a compound's molecular formula?

    <p>To find the number of empirical units per molecule</p> Signup and view all the answers

    If a compound has a molar mass of 100 g/mol and an empirical formula mass of 25 g/mol, how many formula units are in each molecule?

    <p>2</p> Signup and view all the answers

    In determining a compound's molecular formula, what happens when subscripts from the empirical formula are multiplied by n?

    <p>'n' times more atoms are represented</p> Signup and view all the answers

    'Check Your Learning' questions are used for what purpose in educational content like this?

    <p>To test understanding with practice questions</p> Signup and view all the answers

    'Determination of the Molecular Formula for Nicotine' example illustrates which concept?

    <p>'Empirical Formula'</p> Signup and view all the answers

    What is the molecular formula of a compound with a percent composition of 40.03% C, 6.71% H, 53.26% O, and a molecular mass of 120.1 amu?

    <p>C4H6O2</p> Signup and view all the answers

    If a compound has a molar mass of 90 g/mol and an empirical formula mass of 30 g/mol, how many formula units are in each molecule?

    <p>2</p> Signup and view all the answers

    What is the empirical formula of a compound that has a percent composition of 60.00% N, 20.00% H, and 20.00% C?

    <p>NH3</p> Signup and view all the answers

    If a compound's molecular formula is C6H10O5 and its empirical formula mass is 90 g/mol, what is the molar mass of this compound?

    <p>270 g/mol</p> Signup and view all the answers

    Why is it important to compare the molar mass to the empirical formula mass when determining a compound's molecular formula?

    <p>To determine the number of empirical formula units per molecule</p> Signup and view all the answers

    What happens when subscripts from the empirical formula are multiplied by n to obtain the molecular formula?

    <p>The molecular weight increases</p> Signup and view all the answers

    In the context of determining molecular formulas, what does 'n' represent?

    <p>'n' represents the number of empirical formula units per molecule</p> Signup and view all the answers

    Which of these compounds would have a molecular formula represented as (C3H6O3)?

    <p>$C_4H_8O_4$</p> Signup and view all the answers

    'Check Your Learning' questions are primarily used for what purpose in educational content?

    <p>To evaluate student comprehension and understanding</p> Signup and view all the answers

    'Determining the Molecular Formula for Nicotine' example illustrates which concept?

    <p>'Using empirical formulas to find molecular formulas'</p> Signup and view all the answers

    What is the molecular formula of a compound with a percent composition of 49.47% C, 5.201% H, 28.84% N, and 16.48% O, and a molecular mass of 194.2 amu?

    <p>C7H9N5O4</p> Signup and view all the answers

    If a compound has an empirical formula mass of 30 g/mol and a molar mass of 150 g/mol, how many formula units are present in each molecule?

    <p>4</p> Signup and view all the answers

    When determining a compound's molecular formula, what would happen if subscripts from the empirical formula are multiplied by a number greater than the actual value of n?

    <p>The molecular formula will be incorrect.</p> Signup and view all the answers

    In the context of molecular formulas, what does 'n' represent?

    <p>Number of empirical formula units per molecule.</p> Signup and view all the answers

    If a molecule has an empirical formula of C3H5O2 and a molecular mass of 90 amu, what is the molar mass of this compound?

    <p>$90 , \text{g/mol}$</p> Signup and view all the answers

    Why is it important to ensure that the molar ratios of elements in a compound are close to whole numbers when deriving its empirical formula?

    <p>To meet stoichiometric requirements.</p> Signup and view all the answers

    If a compound's molecular mass is 162 g/mol and its empirical formula mass is 18 g/mol, how many formula units are present in each molecule?

    <p>$4$</p> Signup and view all the answers

    When comparing the molar mass to the empirical formula mass in determining a compound's molecular formula, what does a ratio of 2 signify?

    <p>$1$ molecule contains $2$ formula units.</p> Signup and view all the answers

    If a compound has a molar mass of $140 , \text{g/mol}$ and an empirical formula mass of $35 , \text{g/mol}$, what is the empirical formula for this compound?

    <p>$C_3H_3O_3$</p> Signup and view all the answers

    Study Notes

    Percent Composition

    • Percent composition of a compound is the percentage by mass of each element in the compound.
    • It is calculated by dividing the mass of each element by the mass of the compound and multiplying by 100%.
    • The formula to calculate percent composition is:
      • %H = (mass H / mass compound) × 100%
      • %C = (mass C / mass compound) × 100%
      • ...

    Determining Percent Composition from Molecular or Empirical Formulas

    • Percent composition can be derived from the molecular or empirical formula of a compound.
    • The mass of each element in a known mass of compound is needed.
    • The formula to calculate percent composition from a molecular formula is:
      • %C = (number of moles of C × molar mass of C) / molar mass of compound × 100%
      • %H = (number of moles of H × molar mass of H) / molar mass of compound × 100%
      • ...

    Determination of Empirical Formulas

    • Empirical formulas are derived from experimentally measured element masses.
    • The steps to derive an empirical formula are:
      1. Derive the number of moles of each element from its mass.
      2. Divide each element's molar amount by the smallest molar amount to yield subscripts for a tentative empirical formula.
      3. Multiply all coefficients by an integer, if necessary, to ensure that the smallest whole-number ratio of subscripts is obtained.
    • The empirical formula is a symbol representing the relative numbers of a compound's elements.

    Deriving Empirical Formulas from Percent Composition

    • Empirical formulas can be derived from percent composition data.
    • The steps to derive an empirical formula from percent composition are:
      1. Calculate the mass of each element in a convenient mass of compound (e.g., 100 g).
      2. Derive the number of moles of each element from its mass.
      3. Divide each element's molar amount by the smallest molar amount to yield subscripts for a tentative empirical formula.
      4. Multiply all coefficients by an integer, if necessary, to ensure that the smallest whole-number ratio of subscripts is obtained.

    Derivation of Molecular Formulas

    • Molecular formulas are derived from the empirical formula and molecular mass (or molar mass).
    • The steps to derive a molecular formula are:
      1. Calculate the empirical formula mass.
      2. Divide the molecular mass (or molar mass) by the empirical formula mass to yield the number of empirical formula units per molecule (n).
      3. Multiply each subscript in the empirical formula by n to obtain the molecular formula.
    • Molecular formulas are symbols representing the absolute numbers of atoms in a compound's molecules.### Determining the Chemical Formula of a Compound
    • Percent Composition: the percentage by mass of each element in a compound
      • Example: a gaseous compound composed of carbon and hydrogen
        • %H = mass H / mass compound × 100%
        • %C = mass C / mass compound × 100%
      • If analysis of a 10.0-g sample showed it to contain 2.5 g H and 7.5 g C, the percent composition would be 25% H and 75% C

    Calculating Percent Composition

    • Example 3.9: analysis of a 12.04-g sample of a liquid compound composed of carbon, hydrogen, and nitrogen
      • %C = 7.34 g C / 12.04 g compound × 100% = 61.0%
      • %H = 1.85 g H / 12.04 g compound × 100% = 15.4%
      • %N = 2.85 g N / 12.04 g compound × 100% = 23.7%

    Determining Percent Composition from Molecular or Empirical Formulas

    • Example 3.10: aspirin with the molecular formula C9H8O4
      • To calculate the percent composition, consider 1 mol of C9H8O4 and use its molar mass (180.159 g/mol)
      • %C = 9 mol C × molar mass C / molar mass C9H8O4 × 100% = 60.00%
      • %H = 8 mol H × molar mass H / molar mass C9H8O4 × 100% = 4.476%
      • %O = 4 mol O × molar mass O / molar mass C9H8O4 × 100% = 35.52%

    Derivation of Empirical Formulas

    • Empirical Formula: symbols representing the relative numbers of a compound's elements
      • Derived from the masses of its constituent elements
      • The most common approach to determining a compound's chemical formula
    • Example 3.11: a sample of the black mineral hematite (Fe2O3) contains 34.97 g of iron and 15.03 g of oxygen
      • Moles of iron and oxygen: 0.6261 mol Fe and 0.9394 mol O
      • Iron-to-oxygen molar ratio: 1.000 mol Fe to 1.500 mol O
      • Empirical formula: Fe2O3

    Derivation of Molecular Formulas

    • Molecular Formula: derived by comparing the compound's molecular or molar mass to its empirical formula mass
      • If the molecular (or molar) mass of the substance is known, it may be divided by the empirical formula mass to yield the number of empirical formula units per molecule (n)
      • Molecular formula: (AxBy)n = AnxBny
    • Example 3.13: nicotine with a molecular mass of 180 amu
      • Empirical formula mass: 30 amu (sum of atomic masses of C, H, and O)
      • Number of empirical formula units per molecule: 180 amu / 30 amu = 6
      • Molecular formula: (CH2O)6 = C6H12O6### Empirical Formulas
    • Derived from experimentally measured element masses
    • Steps to derive empirical formula:
      • Derive number of moles of each element from its mass
      • Divide each element's molar amount by the smallest molar amount to yield subscripts for a tentative empirical formula
      • Multiply all coefficients by an integer, if necessary, to ensure that the smallest whole-number ratio of subscripts is obtained

    Calculating Moles of Elements

    • Formula: mass of element (g) × (1 mol / molar mass of element (g/mol))
    • Example: 1.71 g C × (1 mol C / 12.01 g C) = 0.142 mol C

    Converting to Whole-Number Ratios

    • Divide each subscript by the smallest subscript to get a whole-number ratio
    • If decimal subscripts remain, multiply all subscripts by an integer to get whole numbers

    Flow Chart for Deriving Empirical Formula

    • Measure mass of elements
    • Calculate moles of elements
    • Divide by smallest number of moles
    • Convert ratio to whole numbers
    • Write empirical formula

    Examples of Deriving Empirical Formulas

    • Hematite (iron oxide): Fe2O3
    • Gas from bacterial fermentation of grain: CO2
    • Compound with percent composition: CH2O

    Molecular Formulas

    • Derived from empirical formula and molecular or molar mass
    • Molecular mass or molar mass is often determined experimentally
    • Empirical formula mass is the sum of average atomic masses of all atoms represented
    • Molecular formula is derived by comparing molecular or molar mass to empirical formula mass

    Deriving Molecular Formulas

    • Formula: molecular or molar mass (amu or g/mol) / empirical formula mass (amu or g/mol) = n formula units/molecule
    • Multiply each subscript in the empirical formula by n to get the molecular formula
    • Example: (CH2O)6 = C6H12O6

    Example of Deriving Molecular Formula

    • Nicotine: C10H14N2

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    Description

    Test your knowledge on calculating the empirical formula of a compound using experimentally derived data and molar masses. Practice converting masses of elements to moles and deriving whole-number ratios to determine the empirical formula.

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