뒤로General Chemistry I: Combustion, Stoichiometry, Solutions, and Reactions – Guided Study
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Q1. Combustion analysis of an 87.12 mg sample of a compound containing only carbon, hydrogen, and oxygen produced 198.05 mg of CO₂ and 81.09 mg of H₂O. What is the empirical formula for the compound?
Background
Topic: Combustion Analysis & Empirical Formula
This question tests your ability to use combustion data to determine the empirical formula of a compound containing C, H, and O. You must relate the masses of CO₂ and H₂O produced to the original amounts of C and H in the sample, and then deduce the amount of O by difference.
Key Terms and Formulas
Empirical formula: The simplest whole-number ratio of atoms in a compound.
Molar mass of CO₂: 44.01 g/mol
Molar mass of H₂O: 18.02 g/mol
Each mole of CO₂ contains 1 mole of C; each mole of H₂O contains 2 moles of H.
Step-by-Step Guidance
Convert the mass of CO₂ produced to moles using its molar mass:
From the moles of CO₂, determine the moles of C in the original sample (1:1 ratio).
Convert the mass of H₂O produced to moles using its molar mass:
From the moles of H₂O, determine the moles of H in the original sample (2:1 ratio).
Calculate the mass of C and H in the sample, subtract from the total sample mass to find the mass of O, and convert that to moles.
Try solving on your own before revealing the answer!
Final Answer: Empirical formula is C₂H₄O
By calculating the moles of C and H from the combustion products and finding the remaining mass as O, the simplest whole-number ratio is 2:4:1 for C:H:O.
Q2. How many grams of carbon dioxide (CO₂) would be formed if 15.0 grams of hexane (C₆H₁₄) were completely burned in excess oxygen?
Background
Topic: Stoichiometry of Combustion Reactions
This question tests your ability to use stoichiometry to relate the mass of a hydrocarbon fuel to the mass of CO₂ produced during complete combustion.
Key Terms and Formulas
Balanced equation for hexane combustion:
Molar mass of C₆H₁₄: 86.18 g/mol
Molar mass of CO₂: 44.01 g/mol
Step-by-Step Guidance
Calculate the moles of hexane in 15.0 g:
Use the stoichiometric ratio from the balanced equation to find moles of CO₂ produced per mole of hexane.
Multiply the moles of hexane by the appropriate ratio to get moles of CO₂.
Convert moles of CO₂ to grams using its molar mass.
Try solving on your own before revealing the answer!
Final Answer: 92.0 g CO₂
Using the stoichiometry of the reaction and the molar masses, 15.0 g of hexane produces approximately 92.0 g of CO₂ when completely combusted.
Q3. A 15.00 g sample of a compound containing only carbon and hydrogen was burned completely in oxygen. The collected products were 45.43 g of CO₂ and 23.25 g of H₂O. What is the empirical formula of the hydrocarbon?
Background
Topic: Combustion Analysis & Empirical Formula (Hydrocarbons)
This question tests your ability to determine the empirical formula of a hydrocarbon using the masses of CO₂ and H₂O produced during combustion.
Key Terms and Formulas
Empirical formula: Simplest whole-number ratio of atoms.
Molar mass of CO₂: 44.01 g/mol
Molar mass of H₂O: 18.02 g/mol
Each mole of CO₂ gives 1 mole of C; each mole of H₂O gives 2 moles of H.
Step-by-Step Guidance
Convert the mass of CO₂ to moles:
From moles of CO₂, determine moles of C in the sample.
Convert the mass of H₂O to moles:
From moles of H₂O, determine moles of H in the sample (2:1 ratio).
Find the simplest whole-number ratio of C to H to write the empirical formula.
Try solving on your own before revealing the answer!
Final Answer: Empirical formula is C₂H₆
After calculating the moles of C and H and reducing to the simplest ratio, the empirical formula is C₂H₆.
Q4. Calculate the molarity of a 0.175 L sugar solution that was prepared with 0.15 moles of sugar.
Background
Topic: Solution Concentration (Molarity)
This question tests your understanding of how to calculate the molarity of a solution given the amount of solute and the volume of solution.
Key Terms and Formulas
Molarity (M):
Step-by-Step Guidance
Identify the number of moles of sugar (solute): 0.15 mol.
Identify the volume of solution in liters: 0.175 L.
Set up the molarity formula:
Try solving on your own before revealing the answer!
Final Answer: 0.857 M
Dividing 0.15 mol by 0.175 L gives a molarity of 0.857 M.
Q5. Determine the volume of a 0.235 M sugar solution that can be prepared with 0.470 moles of sugar.
Background
Topic: Solution Preparation (Molarity and Volume)
This question tests your ability to rearrange the molarity formula to solve for volume, given moles of solute and desired molarity.
Key Terms and Formulas
Molarity (M):
Rearranged:
Step-by-Step Guidance
Identify the moles of sugar: 0.470 mol.
Identify the desired molarity: 0.235 M.
Set up the formula to solve for volume:
Try solving on your own before revealing the answer!
Final Answer: 2.00 L
Dividing 0.470 mol by 0.235 M gives a volume of 2.00 L.
Q6. A salt solution is to be added to a marine aquarium. Calculate the molarity of a salt solution that is prepared by adding water to 18.65 g of NaCl to give a final volume of 250.0 mL.
Background
Topic: Solution Concentration (Molarity)
This question tests your ability to calculate the molarity of a solution given the mass of solute and the final volume.
Key Terms and Formulas
Molar mass of NaCl: 58.44 g/mol
Molarity (M):
Step-by-Step Guidance
Convert 18.65 g NaCl to moles:
Convert 250.0 mL to liters:
Set up the molarity formula:
Try solving on your own before revealing the answer!
Final Answer: 1.28 M
After converting grams to moles and dividing by the volume in liters, the molarity is 1.28 M.
Q7. The solution in the previous question (B.3) was diluted to 1.000 L. Calculate the molar concentration of the final solution.
Background
Topic: Dilution Calculations
This question tests your understanding of how dilution affects concentration, using the relationship between initial and final concentrations and volumes.
Key Terms and Formulas
Initial concentration () and volume (): from previous question.
Final volume (): 1.000 L.
Formula:
Step-by-Step Guidance
Recall the initial concentration from the previous question.
Set up the dilution equation:
Plug in the known values and solve for (final concentration).
Try solving on your own before revealing the answer!
Final Answer: 0.320 M
After applying the dilution formula, the final concentration is 0.320 M.
Q8. Calculate the volume of a 3.15 M NaOH (aq) solution that should be used to prepare 250. mL of 0.150 M NaOH (aq).
Background
Topic: Solution Dilution
This question tests your ability to use the dilution equation to determine the volume of a concentrated solution needed to prepare a more dilute solution.
Key Terms and Formulas
Initial concentration (): 3.15 M
Final concentration (): 0.150 M
Final volume (): 250. mL = 0.250 L
Formula:
Step-by-Step Guidance
Write down the known values: M, M, L.
Set up the dilution equation:
Rearrange to solve for :
Plug in the values to find .
Try solving on your own before revealing the answer!
Final Answer: 0.0119 L (11.9 mL)
Using the dilution formula, you need 11.9 mL of the 3.15 M NaOH solution to prepare 250 mL of 0.150 M NaOH.
Q9. Toxic chemicals in drinking water are usually reported and safety levels specified in units of parts per million (ppm) by mass. What is the molar concentration of arsenic in a water sample that has 1 ppm arsenic (As)? Keep the answer to 3 sig figs.
Background
Topic: Solution Concentration (ppm to Molarity)
This question tests your ability to convert a concentration given in ppm (by mass) to molarity (mol/L).
Key Terms and Formulas
1 ppm = 1 mg solute per 1 kg (or 1 L) of water (assuming density ≈ 1 g/mL)
Molar mass of As: 74.92 g/mol
Molarity:
Step-by-Step Guidance
Assume 1 L of water; 1 ppm = 1 mg As per 1 L.
Convert 1 mg As to grams:
Convert grams of As to moles:
Since this is per 1 L, the molarity is equal to the number of moles calculated.
Try solving on your own before revealing the answer!
Final Answer: 1.33 × 10⁻⁵ M
1 ppm As corresponds to a molarity of 1.33 × 10⁻⁵ mol/L.
Q10. What is the characteristic identifying feature of a dissociation reaction?
Background
Topic: Dissociation Reactions
This question tests your understanding of what happens during a dissociation reaction, particularly for ionic compounds in water.
Key Terms and Concepts
Dissociation: The process by which an ionic compound separates into its constituent ions in solution.
Step-by-Step Guidance
Recall that in a dissociation reaction, an ionic solid dissolves in water and splits into cations and anions.
Think about the chemical equation format:
Identify the key feature: the formation of free ions in aqueous solution.
Try answering on your own before revealing the answer!
Final Answer:
The characteristic feature is the formation of free, hydrated ions in solution from an ionic compound.
Q11. What is the characteristic identifying feature of a precipitation reaction?
Background
Topic: Precipitation Reactions
This question tests your understanding of what defines a precipitation reaction in aqueous solution.
Key Terms and Concepts
Precipitation: The formation of an insoluble solid (precipitate) when two solutions are mixed.
Step-by-Step Guidance
Recall that a precipitation reaction occurs when two soluble ionic compounds are mixed and form an insoluble product.
Think about the chemical equation:
Identify the key feature: the appearance of a solid precipitate.
Try answering on your own before revealing the answer!
Final Answer:
The characteristic feature is the formation of an insoluble solid (precipitate) from ions in solution.
Q12. Use your textbook or other reference resource to find solubility rules for ionic compounds. Ionic compounds are those compounds that dissolve and dissociate to produce ions in solution or precipitate from ions in solution because they are insoluble.
Background
Topic: Solubility Rules
This question tests your knowledge of which ions generally form soluble or insoluble compounds, and the exceptions to these rules.
Key Terms and Concepts
Solubility rules: Guidelines for predicting whether an ionic compound will dissolve in water.
Step-by-Step Guidance
For (a): List cations that generally form soluble compounds (e.g., alkali metals, ammonium).
For (b): List anions that generally form soluble compounds (e.g., nitrate, acetate, most halides).
For (c): List anions that generally form insoluble compounds (e.g., carbonate, phosphate, sulfide).
For (d): Identify exceptions to these rules (e.g., silver halides, sulfates with Ba²⁺, etc.).
Try answering on your own before revealing the answer!
Final Answer:
(a) Soluble cations: Li⁺, Na⁺, K⁺, Rb⁺, Cs⁺, NH₄⁺ (b) Soluble anions: NO₃⁻, C₂H₃O₂⁻ (acetate), Cl⁻, Br⁻, I⁻ (except with Ag⁺, Pb²⁺, Hg₂²⁺) (c) Insoluble anions: CO₃²⁻, PO₄³⁻, S²⁻, OH⁻ (except with alkali metals and NH₄⁺) (d) Exceptions: AgCl, BaSO₄, PbSO₄, CaSO₄ are insoluble; alkali metal and ammonium salts are always soluble.
Q13. For each of the pairs of reactants in a) through c) below:
Write the formulas for the reactants.
Below these formulas write the ions that these reactants produce through dissociation in aqueous solution.
Predict whether a precipitation reaction will occur. Consider the possible compounds produced when the anions and cations of the reactants exchange partners. If any of these is predicted to be insoluble from your solubility rules, then the reaction is a precipitation reaction.
Write the balanced overall ionic reaction using the formulas for the reactants and products. Be sure to include the (states) for each.
Write the balanced net ionic equation, omitting the spectator ions.
Write the balanced molecular equation.
a. lead(II) acetate and sodium carbonate in water
Background
Topic: Precipitation Reactions & Net Ionic Equations
This question tests your ability to predict products, write balanced equations, and identify precipitates using solubility rules.
Key Terms and Formulas
Lead(II) acetate: Pb(C₂H₃O₂)₂
Sodium carbonate: Na₂CO₃
Solubility rules for carbonates and acetates
Step-by-Step Guidance
Write the formulas for the reactants: Pb(C₂H₃O₂)₂ (aq) and Na₂CO₃ (aq).
Write the ions produced upon dissociation: Pb²⁺, 2 C₂H₃O₂⁻, 2 Na⁺, CO₃²⁻.
Predict the possible products by exchanging ions: PbCO₃ and NaC₂H₃O₂.
Use solubility rules to determine if either product is insoluble (precipitate).
Set up the overall ionic and net ionic equations, but stop before balancing and writing the final equations.
Try completing the equations on your own before revealing the answer!
Final Answer:
Pb(C₂H₃O₂)₂ (aq) + Na₂CO₃ (aq) → PbCO₃ (s) + 2 NaC₂H₃O₂ (aq) Net ionic: Pb²⁺ (aq) + CO₃²⁻ (aq) → PbCO₃ (s)
b. calcium chloride and sodium phosphate in water
Background
Topic: Precipitation Reactions & Net Ionic Equations
This question tests your ability to predict products, write balanced equations, and identify precipitates using solubility rules.
Key Terms and Formulas
Calcium chloride: CaCl₂
Sodium phosphate: Na₃PO₄
Solubility rules for phosphates and chlorides
Step-by-Step Guidance
Write the formulas for the reactants: CaCl₂ (aq) and Na₃PO₄ (aq).
Write the ions produced upon dissociation: Ca²⁺, 2 Cl⁻, 3 Na⁺, PO₄³⁻.
Predict the possible products by exchanging ions: Ca₃(PO₄)₂ and NaCl.
Use solubility rules to determine if either product is insoluble (precipitate).
Set up the overall ionic and net ionic equations, but stop before balancing and writing the final equations.
Try completing the equations on your own before revealing the answer!
Final Answer:
3 CaCl₂ (aq) + 2 Na₃PO₄ (aq) → Ca₃(PO₄)₂ (s) + 6 NaCl (aq) Net ionic: 3 Ca²⁺ (aq) + 2 PO₄³⁻ (aq) → Ca₃(PO₄)₂ (s)
c. ammonium chloride and copper (II) sulfate in water
Background
Topic: Precipitation Reactions & Net Ionic Equations
This question tests your ability to predict products, write balanced equations, and identify precipitates using solubility rules.
Key Terms and Formulas
Ammonium chloride: NH₄Cl
Copper(II) sulfate: CuSO₄
Solubility rules for ammonium, chlorides, and sulfates
Step-by-Step Guidance
Write the formulas for the reactants: NH₄Cl (aq) and CuSO₄ (aq).
Write the ions produced upon dissociation: NH₄⁺, Cl⁻, Cu²⁺, SO₄²⁻.
Predict the possible products by exchanging ions: NH₄₂SO₄ and CuCl₂.
Use solubility rules to determine if either product is insoluble (precipitate).
Set up the overall ionic and net ionic equations, but stop before balancing and writing the final equations.
Try completing the equations on your own before revealing the answer!
Final Answer:
2 NH₄Cl (aq) + CuSO₄ (aq) → (NH₄)₂SO₄ (aq) + CuCl₂ (aq) No precipitate forms; all products are soluble, so no net ionic equation for a precipitate.