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- Chemistry: A Study of Matter
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Before viewing an episode, download and print the note-taking guides, worksheets, and lab data sheets for that episode, keeping the printed sheets in order by page number. During the lesson, watch and listen for instructions to take notes, pause the video, complete an assignment, and record lab data. See your classroom teacher for specific instructions.
Support Materials
Chemistry: a study of matter episode.
This semester begins with the introduction of the mole. This important concept will be used during the remainder of the year as the basis for many calculations involving chemical reactions, solutions, and gases. In the units on thermochemistry and chemical kinetics, you will learn how energy is absorbed and given off during chemical reactions and how energy, and factors affect the rates of reactions. The study of reaction rates will lead you into the study of chemical equilibrium. In this semester, you will also study electrochemistry, from batteries to electroplating. And you will learn about acids, bases and salts. Unit 15 is a brief study of the atomic nucleus, which your teacher may introduce at any time during the year.
Chemistry 701: Introduction to the Mole and Molar Mass
Chemistry 702: Percentage Composition and Empirical Formulas
Chemistry 703: Molecular Formulas and Hydrates
Chemistry 802: Mass/Mass Stoichiometry Problems and Percent Yield
Chemistry 803: Limiting Reactants
Chemistry 901: Kinetic Theory, Atmospheric Pressure, and Gas Pressure
Chemistry 902: Boyle's Law and Charles' Law
Chemistry 903: More About the Behavior of Gases
Chemistry 1001: Solutions: A Special Type of Mixture
Chemistry 1002: Solubility
Chemistry 1003: Molarity and Colligative Properties
Chemistry 1101: Introduction to Acids, Bases, and Salts
Chemistry 1102: Indicators and the pH Scale
Chemistry 1103: Neutralization Reactions
Chemistry 1201: Reaction Rates
Chemistry 1202: LeChatelier's Principle, Keq
Chemistry 1301: Thermochemistry
Chemistry 1401: Electrochemistry
Chemistry 1501: Nuclear Science – Part I
Chemistry 1502: Nuclear Science – Part II
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- Chemical Engineering
chemistry worksheet # 2: the mole as a unit of mass
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6.5: Mole-Mass and Mass-Mass Problems
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Skills to Develop
- To convert from mass or moles of one substance to mass or moles of another substance in a chemical reaction.
We have established that a balanced chemical equation is balanced in terms of moles as well as atoms or molecules. We have used balanced equations to set up ratios, now in terms of moles of materials, that we can use as conversion factors to answer stoichiometric questions, such as how many moles of substance A react with so many moles of reactant B. We can extend this technique even further. Recall that we can relate a molar amount to a mass amount using molar mass. We can use that ability to answer stoichiometry questions in terms of the masses of a particular substance, in addition to moles. We do this using the following sequence:
Collectively, these conversions are called mole-mass calculations.
As an example, consider the balanced chemical equation
\[Fe_2O_3 + 3SO_3 \rightarrow Fe_2(SO_4)_3 \label{Eq1}\]
If we have 3.59 mol of Fe 2 O 3 , how many grams of SO 3 can react with it? Using the mole-mass calculation sequence, we can determine the required mass of SO 3 in two steps. First, we construct the appropriate molar ratio, determined from the balanced chemical equation, to calculate the number of moles of SO 3 needed. Then using the molar mass of SO 3 as a conversion factor, we determine the mass that this number of moles of SO 3 has.
The first step resembles the exercises we did in Section 6.4 "Mole-Mole Relationships in Chemical Reactions". As usual, we start with the quantity we were given:
\[\mathrm{3.59\: mol\: Fe_2O_3\times\dfrac{3\: mol\: SO_3}{1\: mol\: Fe_2O_3}=10.77\: mol\: SO_3} \label{Eq2}\]
The mol Fe 2 O 3 units cancel, leaving mol SO 3 unit. Now, we take this answer and convert it to grams of SO 3 , using the molar mass of SO 3 as the conversion factor:
\[\mathrm{10.77\: mol\: SO_3\times\dfrac{80.06\: g\: SO_3}{1\: mol\: SO_3}=862\: g\: SO_3} \label{Eq3}\]
Our final answer is expressed to three significant figures. Thus, in a two-step process, we find that 862 g of SO 3 will react with 3.59 mol of Fe 2 O 3 . Many problems of this type can be answered in this manner.
The same two-step problem can also be worked out in a single line, rather than as two separate steps, as follows:
We get exactly the same answer when combining all the math steps together as we do when we calculate one step at a time.
Example \(\PageIndex{1}\)
How many grams of CO 2 are produced if 2.09 mol of HCl are reacted according to this balanced chemical equation?
\[CaCO_3 + 2HCl \rightarrow CaCl_2 + CO_2 + H_2O\]
Our strategy will be to convert from moles of HCl to moles of CO 2 and then from moles of CO 2 to grams of CO 2 . We will need the molar mass of CO 2 , which is 44.01 g/mol. Performing these two conversions in a single-line gives 46.0 g of CO 2 :
The molar ratio between CO 2 and HCl comes from the balanced chemical equation.
It is a small step from mole-mass calculations to mass-mass calculations. If we start with a known mass of one substance in a chemical reaction (instead of a known number of moles), we can calculate the corresponding masses of other substances in the reaction. The first step in this case is to convert the known mass into moles, using the substance’s molar mass as the conversion factor. Then—and only then—we use the balanced chemical equation to construct a conversion factor to convert that quantity to moles of another substance, which in turn can be converted to a corresponding mass. Sequentially, the process is as follows:
This three-part process can be carried out in three discrete steps or combined into a single calculation that contains three conversion factors. The following example illustrates both techniques.
Example \(\PageIndex{2}\) :Chlorination of Carbon
Methane can react with elemental chlorine to make carbon tetrachloride (CCl 4 ). The balanced chemical equation is as follows:
\[CH_4 + 4Cl_2 \rightarrow CCl_4 + 4HCl\]
How many grams of HCl are produced by the reaction of 100.0 g of CH 4 ?
First, let us work the problem in stepwise fashion. We begin by converting the mass of CH 4 to moles of CH 4 , using the molar mass of CH 4 (16.05 g/mol) as the conversion factor:
\(\mathrm{100.0\: g\: CH_4\times\dfrac{1\: mol\: CH_4}{16.05\: g\: CH_4}=6.231\: mol\: CH_4}\)
Note that we inverted the molar mass so that the gram units cancel, giving us an answer in moles. Next, we use the balanced chemical equation to determine the ratio of moles CH 4 and moles HCl and convert our first result into moles of HCl:
\(\mathrm{6.231\: mol\: CH_4\times\dfrac{4\: mol\: HCl}{1\: mol\: CH_4}=24.92\: mol\: HCl}\)
Finally, we use the molar mass of HCl (36.46 g/mol) as a conversion factor to calculate the mass of 24.92 mol of HCl:
\(\mathrm{24.92\: mol\: HCl\times\dfrac{36.46\: g\: HCl}{1\: mol\: HCl}=908.5\: g\: HCl}\)
In each step, we have limited the answer to the proper number of significant figures. If desired, we can do all three conversions on a single line:
\(\mathrm{100.0\: g\: CH_4\times\dfrac{1\: mol\: CH_4}{16.05\: g\: CH_4}\times\dfrac{4\: mol\: HCl}{1\: mol\: CH_4}\times\dfrac{36.46\: g\: HCl}{1\: mol\: HCl}=908.7\: g\: HCl}\)
This final answer is slightly different from our first answer because only the final answer is restricted to the proper number of significant figures. In the first answer, we limited each intermediate quantity to the proper number of significant figures. As you can see, both answers are essentially the same.
To Your Health: The Synthesis of Taxol
Taxol is a powerful anticancer drug that was originally extracted from the Pacific yew tree ( Taxus brevifolia ). As you can see from the accompanying figure, taxol is a very complicated molecule, with a molecular formula of C 47 H 51 NO 14 . Isolating taxol from its natural source presents certain challenges, mainly that the Pacific yew is a slow-growing tree, and the equivalent of six trees must be harvested to provide enough taxol to treat a single patient. Although related species of yew trees also produce taxol in small amounts, there is significant interest in synthesizing this complex molecule in the laboratory.
After a 20-year effort, two research groups announced the complete laboratory synthesis of taxol in 1994. However, each synthesis required over 30 separate chemical reactions, with an overall efficiency of less than 0.05%. To put this in perspective, to obtain a single 300 mg dose of taxol, you would have to begin with 600 g of starting material. To treat the 26,000 women who are diagnosed with ovarian cancer each year with one dose, almost 16,000 kg (over 17 tons) of starting material must be converted to taxol. Taxol is also used to treat breast cancer, with which 200,000 women in the United States are diagnosed every year. This only increases the amount of starting material needed.
Clearly, there is intense interest in increasing the overall efficiency of the taxol synthesis. An improved synthesis not only will be easier but also will produce less waste materials, which will allow more people to take advantage of this potentially life-saving drug.
Figure \(\PageIndex{1}\) The Structure of the Cancer Drug Taxol. Because of the complexity of the molecule, hydrogen atoms are not shown, but they are present on every atom to give the atom the correct number of covalent bonds (four bonds for each carbon atom).
Concept Review Exercises
- What is the general sequence of conversions for a mole-mass calculation?
- What is the general sequence of conversions for a mass-mass calculation?
- mol first substance → mol second substance → mass second substance
- mass first substance → mol first substance → mol second substance → mass second substance
Key Takeaway
- A balanced chemical equation can be used to relate masses or moles of different substances in a reaction.
Given the following unbalanced chemical equation,
H 3 PO 4 + NaOH → H 2 O + Na 3 PO 4
what mass of H 2 O is produced by the reaction of 2.35 mol of H 3 PO 4 ?
Precipitation reactions, in which a solid (called a precipitate) is a product, are commonly used to remove certain ions from solution. One such reaction is as follows:
Ba(NO 3 ) 2 (aq) + Na 2 SO 4 (aq) → BaSO 4 (s) + 2NaNO 3 (aq)
How many grams of Na 2 SO 4 are needed to precipitate all the barium ions produced by 43.9 g of Ba(NO 3 ) 2 ?
Antacids are bases that neutralize acids in the digestive tract. Magnesium hydroxide [Mg(OH) 2 ] is one such antacid. It reacts with hydrochloric acid in the stomach according to the following reaction:
Mg(OH) 2 + 2HCl → MgCl 2 + 2H 2 O
How many grams of HCl can a two hundred milligram (2.00x10 2 mg) dose of Mg(OH) 2 neutralize?
A simplified version of the processing of iron ore into iron metal is as follows:
2Fe 2 O 3 + 3C → 4Fe + 3CO 2
How many grams of C are needed to produce 1.00 × 10 9 g of Fe?
- 1.61 × 10 8 g
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How many moles of nitrogen gas will be produced if 0.55 moles of oxygen react? 0.371 mol N2. Study with Quizlet and memorize flashcards containing terms like 1. Cu+O2->CuO (balance), If 101 grams of copper is used, how many miles of copper (II) oxide will be formed?, If 5.25 moles of copper are used, how many moles of oxygen must also be used ...
answer The only new concept we will introduce in this unit is the idea of a mole. A mole is a quantity of matter that we use for conversion purposes. We can convert from grams to moles, liters to moles (for gases), and atoms or molecules to moles.
Expert-verified View the full answer Previous question Next question Transcribed image text: CHEMISTRY B-MOLES PACKET NAME: HR: PAGE 9 CHEMISTRY WORKSHEET #6 MIXED MOLE PROBLEMS (GRAMS, MOLECULES, AND LITERS) You now know three things a mole can be: a molar mass, 6.02 x 10 molecules and for a gas, 22.4 liters at STP.
Solution. a) 3.8 x 10 51 grams. b) 4.11 x 10 45 grams. c) 7.31 x 10 25 grams. 4.7.1: Mole Practice with Guidence (Exercises) is shared under a not declared license and was authored, remixed, and/or curated by LibreTexts. These are homework exercises to accompany the Textmap created for Chemistry: A Molecular Approach by Nivaldo Tro.
1 Mass of substance Convert using the molar mass of the substance. 2 Amount of substance in moles Use Avogadro's number for conversion. 3 Number of atoms, molecules, or formula units of substance PROBLEMS INVOLVING ATOMS AND ELEMENTS Sample Problem 1 A chemist has a jar containing 388.2 g of iron filings.
Science > Chemistry library > Chemical reactions and stoichiometry > Stoichiometry Converting moles and mass Google Classroom You might need: Calculator The molecular weight of sodium chloride, NaCl , is 58.44 g mol . How many moles of salt are in 13.8 g of sodium chloride? Express the answer using 3 significant figures. mol Show Calculator Stuck?
Google Classroom You might need: Calculator Using the information in the table, calculate the number of moles in a 2.03 kg sample of citric acid ( C A 6 H A 8 O A 7 ). Write your answer using three significant figures. mol C A 6 H A 8 O A 7 Show Calculator Stuck? Review related articles/videos or use a hint. Report a problem Do 4 problems
CHEMISTRY WORKSHEET # 6 MIXED MOLE PROBLEMS (GRAMS, MOLECULES, AND LITERS) You now know three things a mole can be: a molar mass, 6.02 x 1023 molecules and, for a gas, 22.4 liters at STP. We can use this information to convert grams to molecules or liters, molecules to grams or liters, or liters to grams or molecules.
1. ___Cu + ___O2. ___CuO. a. If 101 grams of copper is used, how many moles of copper (II) oxide will be formed? b. If 5.25 moles of copper are used, how many moles of oxygen must also be used? c. If 78.2 grams of oxygen react with copper, how many moles of copper (II) oxide will be produced?
Our strategy will be to convert from moles of HCl to moles of CO 2 and then from moles of CO 2 to grams of CO 2. We will need the molar mass of CO 2, which is 44.01 g/mol. Performing these two conversions in a single-line gives 46.0 g of CO 2: The molar ratio between CO 2 and HCl comes from the balanced chemical equation.
1. How many atoms are in 6.2 moles of aluminum? 2. Convert 5.3 x 1025 molecules of CO2 to moles. 3. How many formula units of sodium acetate are in 0.87 moles of sodium acetate? 4. Convert 3.55 moles NaCl to formula units. CHEMISTRY: A Study of Matter 2004, GPB 7.5 5. Convert 3.00 moles As2S3 to grams. 6.
a. 2.48 g of HBr. b. 4.77 g of CS 2. c. 1.89 g of NaOH. d. 1.46 g of SrC 2 O 4. 16. Decide whether each statement is true or false and explain your reasoning. There are more molecules in 0.5 mol of Cl 2 than in 0.5 mol of H 2. One mole of H 2 has 6.022 × 10 23 hydrogen atoms. The molecular mass of H 2 O is 18.0 amu.
1. What is the difference between mass and weight? 2. What unit is used to measure mass? Weight? 3. What is the small unit of a covalent compound? An ionic compound? Molar mass tells us the mass ("weight") of 1 mol of an atom or compound.
Answer each of the following questions using the equation provided. BE SURE TO BALANCE EACH EQUATION BEFORE SOLVING ANY PROBLEMS. SHOW ALL WORK. 1. ___Cu 2 + ___O2 ___CuO 2 a. If 101 grams of copper is used, how many moles of copper (II) oxide will be formed? ? mol CuO = 101 g Cu 1 mol Cu 2 mol CuO 63.5 g Cu 2 mol Cu = 1.59 mol CuO b.
Answer. 0.00119 mol C 4 H 8 O 2 which is 0.105 g C 4 H 8 O 2. 7.1.2: Practice Mole Calculations is shared under a not declared license and was authored, remixed, and/or curated by LibreTexts. Back to top. 7.1.1: Chemical Formulas as Conversion Factors.
Directions: For each substance below, state the representative particle (atom, ion, molecule, or formula unit). If the RP is a molecule, state the number of atoms that make up the molecule. If the RP is a formula unit, state the number of ions that make up the formula unit. Formula: # of atoms in the formula: Formula: # of atoms in the formula:
CHEMISTRY WORKSHEET # 7: GENERAL REVIEW OF MOLE PROBLEMS Now we have studied the idea of moles and learned three interpretations of a mole: (1) A gram formula weight. (2) An Avogadro's number of particles. (3) 22.4 liters of gas at STP. Problems 1-2: moles to grams AND grams to moles 1.
Mixed Problems - Mole/Mole and Mole/Mass Worksheet. Challenge Problem: Stoichiometry. This semester begins with the introduction of the mole. This important concept will be used during the remainder of the year as the basis for many calculations involving chemical reactions, solutions, and gases. In the units on thermochemistry and chemical ...
1. Calculate the percent purity of a sample of Mg(OH)2 if titration of 2.568 g of the sample required 38.45 mL of 0.6695 M H3PO4. 3Mg(OH)2 + 2H3PO4 → Mg3(PO4)2 + 6H2O. 2. A 2.00g sample of limestone was dissolved in hydrochloric acid and all the calcium present in the sample was converted to Ca2+ (aq).
CHEMISTRY B- MOLES PACKET NAME: _____ HR: _____ PAGE 9 CHEMISTRY WORKSHEET # 6 MIXED MOLE PROBLEMS (GRAMS, MOLECULES, AND LITERS) You now know three things a mole can be: a molar mass, 6.02 x 1023 molecules and, for a gas, 22.4 liters at STP. We can use this information to convert grams to molecules or liters, molecules to grams or liters, or ...
48.0 molecules NaOH x 1 mole NaOH x 6.022 x 1023 molecules NaOH 40 g NaOH 1 mole NaOH = 7.23 ... x 153.8 g CCl 4 = 1180 g CCl 4 6.022 x 1023 molecules CCl 4 1 mole CCl 4. Title: Mole Calculation Worksheet Author: Moira O'Toole Subject: chemistry Created Date:
The mol Fe 2 O 3 units cancel, leaving mol SO 3 unit. Now, we take this answer and convert it to grams of SO 3, using the molar mass of SO 3 as the conversion factor: (6.5.3) 10.77 m o l S O 3 × 80.06 g S O 3 1 m o l S O 3 = 862 g S O 3. Our final answer is expressed to three significant figures.