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What is the main purpose of calculating percent composition in chemistry?
What is the main purpose of calculating percent composition in chemistry?
In the compound CuBr2, bromine contributes more to the mass percent than copper.
In the compound CuBr2, bromine contributes more to the mass percent than copper.
True
What is the empirical formula for glucose, C6H12O6?
What is the empirical formula for glucose, C6H12O6?
CH2O
The percent composition of nitrogen in (NH4)2S is _________.
The percent composition of nitrogen in (NH4)2S is _________.
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Match the compounds with their percent composition by mass of nitrogen:
Match the compounds with their percent composition by mass of nitrogen:
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Which of the following compounds has the highest percentage of mass of nitrogen?
Which of the following compounds has the highest percentage of mass of nitrogen?
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The calculation of percent composition is only needed for ionic compounds.
The calculation of percent composition is only needed for ionic compounds.
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How do you determine the percent composition of a compound?
How do you determine the percent composition of a compound?
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What is the empirical formula of C6H12O6?
What is the empirical formula of C6H12O6?
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An empirical formula can show the exact number of atoms of each element in a compound.
An empirical formula can show the exact number of atoms of each element in a compound.
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What is the molecular formula of a compound with an empirical formula of C2OH4 and a molecular mass of 88 g/mol?
What is the molecular formula of a compound with an empirical formula of C2OH4 and a molecular mass of 88 g/mol?
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A sample contains 0.3374g of P, 0.0220g of H, and the remainder is O. The empirical formula is __________.
A sample contains 0.3374g of P, 0.0220g of H, and the remainder is O. The empirical formula is __________.
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Match the following compounds with their corresponding empirical formulas:
Match the following compounds with their corresponding empirical formulas:
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How many moles of oxygen (O2) are needed to react with 3.24 moles of ammonia (NH3) in the reaction: _NH3 + _O2 → _NO + _H2O?
How many moles of oxygen (O2) are needed to react with 3.24 moles of ammonia (NH3) in the reaction: _NH3 + _O2 → _NO + _H2O?
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If 145.7 grams of manganese (IV) oxide (MnO2) are reacted, how many grams of hydrochloric acid (HCl) are required to completely react with it?
If 145.7 grams of manganese (IV) oxide (MnO2) are reacted, how many grams of hydrochloric acid (HCl) are required to completely react with it?
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The molar mass of lead (II) nitrate (Pb(NO3)2) is essential for calculating the grams needed to produce a given mass of sodium nitrate (NaNO3).
The molar mass of lead (II) nitrate (Pb(NO3)2) is essential for calculating the grams needed to produce a given mass of sodium nitrate (NaNO3).
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What is the percent yield of the reaction if 75.0 g of P4 yields 111.0 g of PCl3?
What is the percent yield of the reaction if 75.0 g of P4 yields 111.0 g of PCl3?
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In the reaction of Fe(OH)2 with H3PO4, both reactants will yield the same amount of product.
In the reaction of Fe(OH)2 with H3PO4, both reactants will yield the same amount of product.
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What is the limiting reagent when 3.20 g of Fe(OH)2 reacts with 2.50 g of H3PO4?
What is the limiting reagent when 3.20 g of Fe(OH)2 reacts with 2.50 g of H3PO4?
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When octane (C8H18) is burned, _____ and water are produced.
When octane (C8H18) is burned, _____ and water are produced.
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Match the following reactions with their expected yields:
Match the following reactions with their expected yields:
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How many grams of water can be produced from the complete combustion of 320 g of octane?
How many grams of water can be produced from the complete combustion of 320 g of octane?
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If 98.03 g of Magnesium hydroxide decomposes, no water will be produced.
If 98.03 g of Magnesium hydroxide decomposes, no water will be produced.
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How many grams of Fe3(PO4)2 precipitate can be formed from the reaction of Fe(OH)2 and H3PO4?
How many grams of Fe3(PO4)2 precipitate can be formed from the reaction of Fe(OH)2 and H3PO4?
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What is the empirical formula for a compound containing 18.7g of C and 6.28g of H?
What is the empirical formula for a compound containing 18.7g of C and 6.28g of H?
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Multiple compounds can have different empirical formulas.
Multiple compounds can have different empirical formulas.
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What is needed to determine the true formula of a compound aside from the empirical formula?
What is needed to determine the true formula of a compound aside from the empirical formula?
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The ____ reactions tell you how many molecules and/or atoms of each reactant are needed to make the products.
The ____ reactions tell you how many molecules and/or atoms of each reactant are needed to make the products.
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Match the following compounds with their respective formulas:
Match the following compounds with their respective formulas:
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In converting element masses to moles, what can be determined?
In converting element masses to moles, what can be determined?
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The balanced chemical equation for aluminum reacting with oxygen produces aluminum chloride.
The balanced chemical equation for aluminum reacting with oxygen produces aluminum chloride.
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If 24 atoms of aluminum are present, how many molecules of aluminum oxide will be produced?
If 24 atoms of aluminum are present, how many molecules of aluminum oxide will be produced?
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How many moles of calcium carbonate are formed when 0.50 moles of calcium oxide reacts with 0.40 moles of carbon dioxide?
How many moles of calcium carbonate are formed when 0.50 moles of calcium oxide reacts with 0.40 moles of carbon dioxide?
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The theoretical yield of ammonia (NH3) can exceed 10.5 g when 15.0 g of nitrogen gas reacts with hydrogen gas.
The theoretical yield of ammonia (NH3) can exceed 10.5 g when 15.0 g of nitrogen gas reacts with hydrogen gas.
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If 0.996 g of aluminum hydroxide is recovered from a reaction involving 2.80 g of aluminum nitrate, what is the actual yield of aluminum hydroxide?
If 0.996 g of aluminum hydroxide is recovered from a reaction involving 2.80 g of aluminum nitrate, what is the actual yield of aluminum hydroxide?
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To produce 30.0 g of aluminum phosphate with a % yield of 73.7%, the theoretical mass of aluminum needed is _____ g.
To produce 30.0 g of aluminum phosphate with a % yield of 73.7%, the theoretical mass of aluminum needed is _____ g.
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What information is crucial when calculating the percent yield of a chemical reaction?
What information is crucial when calculating the percent yield of a chemical reaction?
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Match the chemical reaction types with their appropriate description:
Match the chemical reaction types with their appropriate description:
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The limiting reactant in a chemical reaction is always the reactant with the smallest amount in moles.
The limiting reactant in a chemical reaction is always the reactant with the smallest amount in moles.
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What do you calculate first when determining the mass of reactants needed to produce a desired mass of products?
What do you calculate first when determining the mass of reactants needed to produce a desired mass of products?
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What is the balanced chemical equation for the reaction of nitrogen and hydrogen to form ammonia?
What is the balanced chemical equation for the reaction of nitrogen and hydrogen to form ammonia?
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The limiting reactant is the reactant that is left over after a reaction has taken place.
The limiting reactant is the reactant that is left over after a reaction has taken place.
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What is percent yield?
What is percent yield?
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In a chemical reaction, the full amount of product produced is called __________ yield.
In a chemical reaction, the full amount of product produced is called __________ yield.
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If a student obtained 6.5 g of NH3 from the reaction of 6.0 g of N2 and 18.0 g of H2, what is the formula to calculate the percent yield?
If a student obtained 6.5 g of NH3 from the reaction of 6.0 g of N2 and 18.0 g of H2, what is the formula to calculate the percent yield?
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In the reaction of nitrogen and hydrogen, how do you determine the limiting reactant using masses?
In the reaction of nitrogen and hydrogen, how do you determine the limiting reactant using masses?
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Match the following terms with their definitions:
Match the following terms with their definitions:
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The percent yield can be calculated using the formula: Percent yield = __________ yield / Theoretical yield * 100.
The percent yield can be calculated using the formula: Percent yield = __________ yield / Theoretical yield * 100.
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Study Notes
Unit 7: Stoichiometry
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Learning Objectives: Apply first-semester learning to convert word equations into balanced chemical equations. Include identifying ionic and covalent compounds, diatomic elements, polyatomic ions, and naming/balancing chemical equations.
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Percent Composition: Calculate the percent composition of each element in a molecule.
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Empirical Formula: Calculate the empirical formula of a substance using mass quantities or percent composition.
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Molecular Formula: Calculate the molecular formula using mass quantities, percent composition, and molar mass.
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Mole Ratios: Given balanced chemical equations, determine the theoretical yield and find the limiting reagent.
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Limiting Reactant: Identify the reactant fully consumed in a reaction.
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Excess Reactant: Determine the amount of reactant left over when a reaction has reached completion.
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Percent Yield:
- Calculate experimentally determined actual yield vs theoretical yield
- Measure percentage using the ratio of actual yield to theoretical yield and 100 as the denominator.
Worksheet Contents
- WS#1: Percent Composition
- WS#2: Percentage Composition and Empirical Formula
- WS#3: Reaction Stoichiometry
- WS#4: Limiting and Excess Reactant Stoichiometry
- WS#5: Limiting Reactant and Percent Yield Practice
- WS#6: Word Equations and Stoichiometry
- U7 Lecture Notes: (Pages 12-18- not graded)
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Description
Test your knowledge on the calculation of percent composition and empirical formulas in chemistry. This quiz includes questions on compounds such as CuBr2 and glucose, and challenges you to match compounds with their corresponding mass percent of nitrogen. Dive deep into the essential concepts of chemistry related to molecular and empirical formulas.