Grams, Moles, Molecules: Self-Study Assignment

Grams, Moles, Molecules: Self-Study Assignment You will have a QUIZ on the attached pages on. Your assignment is: READ the pages attached. WORK the examples in the lesson. Complete the pages as homework. To work the examples, use a sheet of paper to cover below the line, try the problem on your paper, then check your answer below the line. Start early. This assignment will require 2-4 hours of work outside of class. 2008 ChemReview.net v. e4 Page vii

Module 8 Grams and Moles Timing: Start this module when you are assigned problems using moles. Prerequisites: You need to have completed Modules 2, 4, 5, and Lessons 6A and 6C. The other lessons are helpful, but not essential, for Module 8. Lesson 8A: The Mole Pretest: With a perfect score on this pretest, you may skip to Lesson 8B. Answers are at the end of this lesson. 1. How many molecules are in a) 4 moles? b) Half a mole? 2. Why is it convenient to pick our unit for counting molecules to be 6.02 x 10 23? Counting Particles Molecules are extremely small. Visible quantities of a substance must therefore have a very large number of molecules, atoms, or ions. One drop of water, for example, contains about 1,500,000,000,000,000,000,000 (1.5 x 10 21 ) water molecules. Rather than writing numbers of this size when solving chemistry calculations, chemists use a convenient unit to count particles. As we count eggs by the dozen, or buy printer paper by the ream (500 sheets), we count chemical particles such as molecules, atoms, and ions by the mole. A mole is 6.02 x 10 23 particles. The number 6.02 x 10 23 is called Avogadro s number. The mole was originally defined by counting the number of atoms in one gram of hydrogen, the lightest element. The definition has changed slightly over time. We now base our count on certain atoms of carbon, but the original definition based on hydrogen remains close to true. Picking a number for the mole, our counting unit, so that 1 mole of the lightest element (hydrogen) has a mass close to one gram, simplifies the arithmetic in problems with other elements, especially when we convert between grams and moles. That s our goal: to calculate a count of particles by measuring their mass on a balance or scale. Counting the particles will make chemical reactions easier to understand. The mole is therefore the most important unit in chemistry. In problems, if we do not know the moles involved, the rule is: find moles first. In the SI (official metric) unit system, mole is abbreviated mol. As with all metric abbreviations, the mol abbreviation is not followed by a period, and no distinction is made between singular and plural when the abbreviation is used. 2008 For additional help, visit www.chemreview.net v. m2 Page 142

Working with Moles Working with very large numbers is simplified by using exponential notation (see Module 1). Recall that 1. 10 23 means a 1 followed by 23 zeros: 100,000,000,000,000,000,000,000. 2. When multiplying a number times a number times an exponential, the numbers multiply by the normal rules of arithmetic, but the exponential does not change. Examples: Half a mole = 1/2 x (6.02 x 10 23 ) = 3.01 x 10 23 particles Ten moles = 10 x (6.02 x 10 23 ) = 60.2 x 10 23 = 6.02 x 10 24 particles 0.20 moles = 0.20 x (6.02 x 10 23 ) = 1.2 x 10 23 particles Practice: Cover the answers at the bottom of this page. (Sticky notes make good answer covers). Write then check your answers below. 1. How many particles are in a. 2.0 moles? b. One third of a mole? c. 5.0 moles? 2. Why is it convenient to set the value for our counting unit at 6.02 x 10 23? ANSWERS Pretest 1a. 24 x 10 23 1b. 3.0 x 10 23 2. For hydrogen, the lightest element, one gram will roughly equal the mass of one mole of H atoms. Practice 1a) 12 x 10 23 particles 1b) 2.0 x 10 23 particles 1c) 30. x 10 23 = 3.0 x 10 24 particles 2a) For the lightest element (H), one gram approximately equals the mass of one mole of H atoms. 2008 For additional help, visit www.chemreview.net v. m2 Page 143

Lesson 8B: Grams Per Mole (Molar Mass) Pretest: Answer this question correctly and you may skip to Lesson 8C. The answer is at the end of this lesson. Q. 1 mole (NH 4 ) 2 SO 4 =? grams (NH 4 ) 2 SO 4 Atomic Mass Each element has a different average mass. The average atom of carbon has a mass approximately 12 times that of the average hydrogen, because carbon has more protons, neutrons, and electrons. Uranium atoms have on average about 238 times more mass than hydrogen atoms. The average mass of an atom is its atomic mass. Atomic mass is measured in atomic mass units (see Lesson 6B). Average atomic masses for the atoms are listed at the end of this booklet and inside the cover of most chemistry textbooks. To encourage mental arithmetic, the atomic masses in these lessons use fewer significant figures than most textbooks. If you use a textbook table of atomic masses, your answers will differ slightly from the answers shown here. Molar Mass The mass of a mole of atoms is the molar mass of the atom. The number that represents the average atomic mass of an atom is the same as the number that measures the molar mass of an atom in units of grams per one mole. The molar mass of the lightest atom, hydrogen, is 1.008 g/mol. A mole of uranium atoms has a mass of 238 grams. For substances that contain more than one atom, the molar mass is easily determined. Simply add the molar masses of each of the atoms that make up the particle. Example: What is the molar mass of NaOH? Add these molar masses: Na = 23.0 O = 16.0 H = 1.008 40.008 = 40.0 g/mol NaOH SF: Recall from Lesson 3A that when adding significant figures, because the highest place with doubt in the columns is in the tenth s place, the sum has doubt in the tenth s place. Round the answer to that place. The molar mass supplies an equality: 40.0 grams NaOH = 1 mole NaOH. After calculating a molar mass, the equality format should be written in the DATA. Include the formula for the substance on both sides of the equality. This will greatly simplify the chemical reaction calculations in upcoming lessons. Molar Masses and Subscripts To calculate the molar mass from chemical formulas containing subscripts, recall that subscripts are exact numbers. Multiplying by a subscript therefore does not change the doubtful digit s place in the result. 2008 For additional help, visit www.chemreview.net v. m2 Page 144

When calculating a molar mass, use the following column format to keep track of the numbers and the place with doubt. Example: Find the molar mass of phosphoric acid: H 3 PO 4 1 mol H 3 PO 4 = 3 mol H = 3 x 1.008 g/mol = 3.024 1 mol P = 1 x 31.0 g/mol = 31.0 4 mol O = 4 x 16.0 g/mol = 64.0 98.024 98.0 g/mol In your DATA, write: 98.0 g H 3 PO 4 = 1 mol H 3 PO 4 The Importance of Molar Mass If you know the substance formula, you can calculate the molar mass of the substance, and the molar mass will convert between mass and moles. The molar mass is probably the most frequently used conversion in chemistry. Why? Grams are easy to measure: use a scale. Moles are difficult to measure directly, since we don t have machines that count large numbers of particles. But moles is the important unit in chemistry, because moles is the unit that counts of visible numbers of particles, and most chemical processes are explained by counting particles. The molar mass allows us to convert between the grams we can measure and the particle counts that explain chemistry. Practice To speed your progress, try the last letter of each problem. If you have difficulty, try other letters of the same problem. Answers are at the end of this lesson. 1. Use your table of elements to find the molar mass of these elements. Include the unit with your answer. a. Nitrogen b. Au c. Pb 2. How many oxygens are represented in each of these formulas? a. Ca(OH) 2 b. Al 2 (SO 4 ) 3 c. Co(NO 3 ) 2 Do the next two in your notebook. Allow yourself enough room on the paper for clear and careful work. Your calculations should look like the example above. 3. Calculate the molar mass for these compounds. Include the units of the answer and proper sf. a. H 2 b. NaH c. KSCN d. Na 3 PO 4 e. Barium nitrate 4. a. 1 mole H 2 S =? grams H 2 S b.? grams AgNO 3 = 1 mole AgNO 3 2008 For additional help, visit www.chemreview.net v. m2 Page 145

ANSWERS Pretest: 1 mole (NH 4 ) 2 SO 4 = 132.2 grams (NH 4 ) 2 SO 4 Practice 1a. Nitrogen 14.0 grams b. Au 197.0 g c. Pb 207.2 g mole mol mol 2. a. Ca(OH) 2 2 oxygens b. Al 2 (SO 4 ) 3 12 oxygens c. Co(NO 3 ) 2 6 oxygens 3a. H 2 = 2 x H = 2 x 1.008 = 2.016 g/mol In your DATA, write 2.016 g H 2 = 1 mol H 2 (Multiplying by an exact subscript does not change the place with doubt). 3b. NaH = 3c. KSCN = Na = 23.0 K = 39.1 H = 1.008 S = 32.1 24.008 = 24.0 g/mol C = 12.0 N = 14.0 In the DATA, write 97.2 g/mol 24.0 grams NaH = 1 mole NaH 97.2 g KSCN = 1 mol KSCN 3d. Na 3 PO 4 = 3e. Barium nitrate = Ba(NO 3 ) 2 (Lesson 7C) = 3 x Na = 3 x 23.0 = 69.0 1 x Ba = 1 x 137.3 = 137.3 1 x P = 1 x 31.0 = 31.0 2 x N = 2 x 14.0 = 28.0 4 x O = 4 x 16.0 = 64.0 6 x O = 6 x 16.0 = 96.0 164.0 g/mol 261.3 g/mol 164.0 g Na 3 PO 4 = 1 mol Na 3 PO 4 261.3 g Ba(NO 3 ) 2 = 1 mol Ba(NO 3 ) 2 4. Question 4 is asking for the grams per one mole. That s the molar mass. 4a. H 2 S = 4b. AgNO 3 = 1 x Ag = 1 x 107.9 = 107.9 2 x H = 2 x 1.008 = 2.016 1 x N = 1 x 14.0 = 14.0 1 x S = 1 x 32.1 = 32.1 3 x O = 3 x 16.0 = 48.0 34.116 = 34.1 g/mol 169.9 g/mol 1 mol H 2 S = 34.1 g H 2 S 169.9 g AgNO 3 = 1 mol AgNO 3 Lesson 8C: Converting Between Grams and Moles Knowing how to calculate the grams per one mole, we now want to be able to calculate the mass of any number of moles of a substance. The problem can be viewed as converting units, in this case from moles to grams. An equality, the molar mass, provides the needed conversion factor. 2008 For additional help, visit www.chemreview.net v. m2 Page 146

The Grams Prompt In theater, a prompt is a word or two that reminds the players of what to do next. In chemistry, certain words or conditions in problems can prompt us to do things automatically that will help in solving problems. Memorize the following rule as the Grams Prompt In a problem, if either in the DATA or WANTED you see grams or prefix-grams (such as kg, mg, etc.) or mass of a substance formula, calculate the molar mass of that formula; then write that molar mass in your DATA as an equality. Example: 1 mole H 2 O = 18.0 grams H 2 O The grams prompt will help to get into your DATA the conversions needed to SOLVE. If you see grams of a formula in a calculation, you will nearly always need the molar mass. Use the grams prompt to solve the following problem. Use the WANTED, DATA, SOLVE format, and then check the answer on the following page. Q. What is the mass in grams of 0.25 moles of O 2? Answer WANTED:? g O 2 = DATA: 0.25 mol O 2 Prompt: (The grams O 2 in the WANTED is a grams prompt. Write the O2 molar mass equality in the DATA. Use the equality as a conversion to solve.) 32.0 g O 2 = 1 mol O 2 SOLVE:? g O 2 = 0.25 mol O 2 32.0 g O 2 = 8.0 g O 2 1 mol O 2 We WANT a single unit (grams), so our data contains a single unit, and it is the given quantity. The other measurements in the DATA will be in pairs, written as equalities or ratios. To convert between grams and moles of a substance, use the molar mass as a conversion factor. SF: Since 0.25 has 2 sf, 32.0 has 3 sf, and 1 is exact, the multiplied answer must be rounded to 2 sf. 2008 For additional help, visit www.chemreview.net v. m2 Page 147

In WANTED, DATA, and conversions, you must write the number, unit, and chemical formula for all terms. By writing the WANTED unit in the problem above, you were prompted to write a conversion that is needed to solve. By listing needed molar masses in the DATA before you start your conversions, you can focus on arranging your conversions properly when you SOLVE. Writing out the WANTED, DATA, prompts, and labels takes time. However, this structured method of problem-solving will greatly improve your success in the complex problems that soon will be encountered. Practice A Try the last letter on each numbered question. If you get it right, go to the next last. If you need more practice to feel confident, do another letter of the numbered problem. Answers are at the end of this lesson. Needed molar masses for problems 2-4 are found in Problem 1 or in the Lesson 8B practice. 1. Working in your notebook, find the molar mass for a. H 2 SO 4 b. Aluminum nitrate 2. Solve.? grams NaOH = 3.5 moles NaOH 40.0 g NaOH = 1 mol NaOH 3. Supply the needed conversion and solve. a.? grams H 2 SO 4 = 4.5 moles H 2 SO 4 b.? g AgNO 3 = 0.050 mol AgNO 3 (For molar mass, see Lesson 8B, Problem 4b.) 4. Use WANTED, DATA, PROMPT and SOLVE to do these in your notebook. a. 2.7 moles of H 2 SO 4 would have a mass of how many grams? b. What would be the mass in grams of 2.0 x 10 6 moles of Al(NO 3 ) 3? (Answer in scientific notation.) Converting Grams to Moles If the grams of a substance with a known formula are known, how do we find the moles? By use the molar mass as a conversion factor to solve. Try the following problem in your notebook. Then check the answer below. Q. How many moles are in 4.00 grams of O 2? 2008 For additional help, visit www.chemreview.net v. m2 Page 148

A. WANT:? mol O 2 = DATA: 4.00 g O 2 SOLVE: 32.0 g O 2 = 1 mol O 2 (grams prompt)? mol O 2 = 4.00 g O 2 1 mol O 2 = 0.125 mol O 2 32.0 g O 2 SF: 4.00 has 3 sf, 1 is exact (infinite sf), 32.0 has 3 sf; answer must be rounded to 3 sf. Practice B Start with the last letter on each numbered question. If you get it right, go to the next number. Need more practice? Do another part. Molar masses for these problems were calculated in the two prior sets of practice. 1. Supply conversions and solve. a.? mol H 2 SO 4 = 10.0 g H 2 SO 4 b.? mol Ba(NO 3 ) 2 = 65.4 g Ba(NO 3 ) 2 2. Solve these in your notebook. a. 19.62 kg of H 2 SO 4 is how many moles? b. How many moles are in 17.0 mg of AgNO 3? (Answer in scientific notation.) ANSWERS Practice A 1a. H 2 SO 4 = 1b. Aluminum nitrate = Al(NO 3 ) 3 (Lesson 7B, 7C) (If needed, adjust your work and finish) 2 x H = 2 x 1.008 = 2.016 1 x Al = 1 x 27.0 = 27.0 1 x S = 1 x 32.1 = 32.1 3 x N = 3 x 14.0 = 42.0 4 x O = 4 x 16.0 = 64.0 9 x O = 9 x 16.0 = 144.0 98.1 g/mol 213.0 g/mol 98.1 g H 2 SO 4 = 1 mol H 2 SO 4 213.0 g Al(NO 3 ) 3 = 1 mol Al(NO 3 ) 3 Note in 1b, for the 9 oxygens, that multiplying by an exact 9 is the same as adding 16.0 nine times. Multiplying by an exact number does not change the place with doubt. 2008 For additional help, visit www.chemreview.net v. m2 Page 149

2.? grams NaOH = 3.5 moles NaOH 40.0 g NaOH = 140 grams NaOH 1 mole NaOH 3a.? grams H 2 SO 4 = 4.5 moles H 2 SO 4 98.1 g H 2 SO 4 = 440 g H 2 SO 4 1 mol H 2 SO 4 3b.? g AgNO 3 = 0.050 moles AgNO 3 169.9 g AgNO 3 = 8.5 g AgNO 3 1 mol AgNO 3 4a.? g H 2 SO 4 = 2.7 mol H 2 SO 4 98.1 g H 2 SO 4 = 260 g H 2 SO 4 1 mol H 2 SO 4 4b.? g Al(NO 3 ) 3 = 2.0 x 10 6 mol Al(NO 3 ) 3 213.0 g Al(NO 3 ) 3 = 4.3 x 10 4 g Al(NO 3 ) 3 1 mol Al(NO 3 ) 3 Practice B 1. a.? mol H 2 SO 4 = 10.0 g H 2 SO 4 1 mol H 2 SO 4 = 0.102 mole H2 SO 4 98.1 g H 2 SO 4 b.? mol Ba(NO 3 ) 2 = 65.4 g Ba(NO 3 ) 2 1 mole Ba(NO 3 ) 2 = 261.3 g Ba(NO 3 ) 2 Answer: If you wrote.250 moles Ba(NO 3 ) 2, mark it wrong. If you wrote 0.250 moles Ba(NO 3 ) 2, go to the head of the class. Always write a 0 in front of a decimal point, if there is no number in front of the decimal point. This makes the decimal point visible when you need this answer for a later step of a lab report or test. 2a.? mol H 2 SO 4 = 19.62 kg H 2 SO 4 10 3 g 1 mol H 2 SO 4 = 200. mol H 2 SO 4 1 kg 98.1 g H 2 SO 4 2b.? mol AgNO 3 = 17.0 mg AgNO 3 1 g 1 mol AgNO 3 = 1.00 x 10 4 mol AgNO 3 10 3 mg 169.9 g AgNO 3 Lesson 8D: Converting Particles, Moles, and Grams Prerequisites: To do this lesson, you need to have completed Lessons 1A and 1B on exponential notation, plus Lessons 8B and 8C. Pretest: If you can solve the following problem, you may skip this lesson. The answer is at the end of the lesson. Q. 2.0 x 10 18 molecules of water have a mass of how many micrograms? 2008 For additional help, visit www.chemreview.net v. m2 Page 150

Problems Involving a Large Number of Particles We know that one mole of anything = 6.02 x 10 23 of anything. That s the definition of a mole. It s like a dozen, only bigger. You will need Avogadro s number in calculations that convert between a count of very small particles (such as molecules, atoms, or ions) and units used to measure visible amounts of particles, such as grams or liters, or when DATA includes a very large exponential term, such as 10 23 or any other two-digit exponent. To identify and solve those calculations, use the Avogadro Prompt If a calculation includes a large exponential (10 xx ) number of particles (molecules, atoms, ions, or ion combinations), or converts between a number of very small particles and base units that measure visible amounts, write under DATA: 1 mole (substance formula) = 6.02 x 10 23 (substance formula) Add that rule to the previous Grams Prompt If the WANTED or DATA includes grams (or prefix-grams) of a substance formula, write in your DATA the molar mass equality for that formula. Example: 1 mole H 2 O = 18.0 grams H 2 O Use the above two prompts to solve the following problem. Use the WANTED, DATA, prompt, and SOLVE format in your notebook, then check your answer below. Q. What is the mass in grams of 1.5 x 10 22 molecules of H 2 O? Answer SOLVE: WANTED:? g H 2 O = DATA: Prompts: 1.5 x 10 22 H 2 O molecules (The 10 xx in the DATA calls the Avogadro prompt.) 1 mol H 2 O = 6.02 x 10 23 H 2 O molecules 1 mol H 2 O = 18.0 g H 2 O (WANTED unit calls the g prompt.)? g H 2 O = 1.5 x 10 22 H 2 O 1 mole H 2 O 18.0 g H 2 O = 6.02 x 10 23 H 2 O 1 mol H 2 O 2008 For additional help, visit www.chemreview.net v. m2 Page 151

= 1.5 x 10 22 18.0 = 4.5 x 10 1 g H 2 O or 0.45 g H 2 O 6.02 x 10 23 There are several ways to do the arithmetic in the problem above. You may use any that works, but try doing the exponential part of the math without the calculator (see Lesson 1C). Showing some steps in the math will often get you partial credit if you make a silly mistake. A rule that many instructors use when grading is this: if it is easy to spot a mistake, take off a little. If you can t understand what was done, take off a lot. On the job, someone will review your work. If you make it easy for your reviewer to see what you have done, so that the occasional mistake is caught before it is shipped out the door, your career will be more rewarding in all respects. Flashcards Add these to your collection. Run them once to perfection, then use them to do the problems below. Repeat for two more days, then put these cards in stack #2 (see Lesson 6E). One-way cards (with notch) To find molar mass of a substance formula The units of molar mass To convert between grams and moles In DATA, write the molar mass as If you see grams or prefix-grams in WANTED or DATA If a calculation includes 10 xx of a substance If a calculation mixes units measuring visible amounts (g, mol, ml ) with units measuring invisibles (atoms, molecules, particles ) Back Side -- Answers Add the molar masses of its atoms Grams per 1 mole Use the molar mass equality 1 mol formula = # g formula Write the molar mass equality in the DATA In the DATA, write 1 mol (formula) = 6.02 x 10 23 (formula) In the DATA, write 1 mol substance = 6.02 x 10 23 of substance Practice Run the flashcards above until you get them right every time. Then solve these problems below in your notebook. Save one problem for your next study session. Answers are on the next page. Problems 3 and 4 are more challenging. 1. 3.55 grams of Cl 2 gas (chlorine gas) contain how many molecules of Cl 2? 2. 8.0 x 10 24 molecules of CH 4 have a mass of how many kilograms? 3. 2.57 nanograms of S 8 would contain how many sulfur atoms? 4. A magnesium ribbon of uniform width and thickness has a mass of 0.750 g Mg/meter. How many magnesium atoms are in 5.25 cm of the ribbon? 2008 For additional help, visit www.chemreview.net v. m2 Page 152

ANSWERS Pretest: 60. μg H 2 O Practice Your paper should look like this, but you may omit the (comments) in parentheses. 1. WANTED:? molecules Cl 2 DATA: 3.55 g Cl 2 71.0 g Cl 2 = 1 mol Cl 2 (grams prompt) 1 mol Cl 2 = 6.02 x 10 23 molecules Cl 2 (mix g and molecules = Avogadro prompt)? molecules Cl 2 = 3.55 g Cl 2 1 mol Cl 2 6.02 x 10 23 molecules Cl 2 = 3.01 x 10 22 71.0 g Cl 2 1 mol Cl 2 molecules Cl 2 2. WANTED:? kg CH 4 DATA: 8.0 x 10 24 CH 4 molecules 1 mol CH 4 = 6.02 x 10 23 CH 4 molecules (10 xx = Avogadro prompt) 1 mol CH 4 = 16.0 g CH 4 (any prefix-grams = grams prompt) SOLVE:? kg CH 4 = 8.0 x 10 24 CH 4 1 mol CH 4 16.0 g CH 4 1 kg = 0.21 kg CH 4 6.02 x 10 23 CH 4 1 mol CH 4 10 3 g 3. WANTED:? atoms S DATA: 2.57 nanograms S 8 1 ng = 10 9 g (listing the less frequently used prefix equalities is a good idea) 256.8 g S 8 = 1 mol S 8 molecules (any prefix-grams = grams prompt) 1 mol S 8 = 6.02 x 10 23 molecules S 8 (both atoms and grams = Avogadro prompt) 1 molecule S 8 = exactly 8 atoms S (Note that the molar mass is the mass of a mole of molecules. A needed conversion is the relationship between molecules and S atoms for S 8. If needed, adjust your work and then complete the problem.)? atoms S = 2.57 ng S 8 10 9 g 1 mol S 8 6.02 x 10 23 molec. S 8 8 atoms S = 4.82 x 10 13 1 ng 256.8 g S 8 1 mol S 8 1 molec. S 8 atoms S (The 8 atoms per molecule is exact, and the 8 therefore does not restrict the number of sf in the answer. Exact numbers have infinite sf.) 2008 For additional help, visit www.chemreview.net v. m2 Page 153

4. WANTED:? atoms Mg DATA: 0.750 g Mg = 1 meter Mg 5.25 cm Mg 24.3 g Mg = 1 mol Mg (grams prompt) SOLVE: 1 mol = 6.02 x 10 23 atoms (atoms = particles = Avogadro prompt) (Want a single unit? Start with single unit.)? atoms Mg = 5.25 cm Mg 1 meter 0.750 g Mg 1 mol Mg 6.02 x 1023 atoms = 9.7 x 10 20 100 cm 1 meter Mg 24.3 g Mg 1 mol atoms Mg SUMMARY: Module 8 Grams and Moles 1. Chemical processes are easiest to understand if you count the particles involved. Large numbers of particles are counted by the mole. 2. 1 mole of anything = 6.02 x 10 23 anythings. That s Avogadro s number. 3. If you know the chemical formula for a substance, you can calculate the grams per mole of the substance, the molar mass. 4. To find the molar mass of a particle, add the molar masses of its atoms. 5. The units of molar mass are grams per one mole. 6. If molar mass is WANTED, write WANTED:? g mol 7. If a molar mass is DATA, write 1 mol formula = XX g formula 8. Grams Prompt If a calculation mentions grams or prefix-grams of a formula, write in your DATA, (Molar Mass) grams formula = 1 mole formula Example: 18.0 g H 2 O = 1 mol H 2 O 9. Avogadro Prompt If a calculation includes 10 xx of a substance, or includes both a count of invisibly particles and base units used to measure visible amounts, write in your DATA: 1 mole (formula) = 6.02 x 10 23 (formula) 10. In most chemistry calculations, the rule to solve will be: find moles first. # # # # # 2008 For additional help, visit www.chemreview.net v. m2 Page 154

Module 10 Stoichiometry NOTE on the Table of Elements The atomic masses in the Table of Elements in the following table use fewer significant figures than most similar tables in college textbooks. By keeping the numbers simple, it is hoped that you will use mental arithmetic to do easy numeric cancellations and simplifications before you use a calculator for arithmetic. Many of the calculations in these lessons have been set up so that you should not need a calculator at all to solve, if you look for easy cancellations first. The purpose is to give you practice at estimating answers. After any use of a calculator, it is important to estimate the answer using mental arithmetic and simple cancellations in order to catch errors in using the calculator. # # # # # 2008 For additional help, visit www.chemreview.net v. m2 Page 200

The ELEMENTS The third column shows the atomic number: The protons in the nucleus of the atom. The fourth column is the molar mass, in grams/mole. For radioactive atoms, ( ) is the molar mass of most stable isotope. Actinium Ac 89 (227) Aluminum Al 13 27.0 Americium Am 95 (243) Antimony Sb 51 121.8 Argon Ar 18 39.95 Arsenic As 33 74.9 Astatine At 84 (210) Barium Ba 56 137.3 Berkelium Bk 97 (247) Beryllium Be 4 9.01 Bismuth Bi 83 209.0 Boron B 5 10.8 Bromine Br 35 79.9 Cadmium Cd 48 112.4 Calcium Ca 20 40.1 Californium Cf 98 (249) Carbon C 6 12.0 Cerium Ce 58 140.1 Cesium Cs 55 132.9 Chlorine Cl 17 35.5 Chromium Cr 24 52.0 Cobalt Co 27 58.9 Copper Cu 29 63.5 Curium Cm 96 (247) Dysprosium Dy 66 162.5 Erbium Er 68 167.3 Europium Eu 63 152.0 Fermium Fm 100 (253) Fluorine F 9 19.0 Francium Fr 87 (223) Gadolinium Gd 64 157.3 Gallium Ga 31 69.7 Germanium Ge 32 72.6 Gold Au 79 197.0 Hafnium Hf 72 178.5 Helium He 2 4.00 Holmium Ho 67 164.9 Hydrogen H 1 1.008 Indium In 49 114.8 Iodine I 53 126.9 Iridium Ir 77 192.2 Iron Fe 26 55.8 Krypton Kr 36 83.8 Lanthanum La 57 138.9 Lawrencium Lr 103 (257) Lead Pb 82 207.2 Lithium Li 3 6.94 Lutetium Lu 71 175.0 Magnesium Mg 12 24.3 Manganese Mn 25 54.9 Mendelevium Md 101 (256) Mercury Hg 80 200.6 Molybdenum Mo 42 95.9 Neodymium Nd 60 144.2 Neon Ne 10 20.2 Neptunium Np 93 (237) Nickel Ni 28 58.7 Niobium Nb 41 92.9 Nitrogen N 7 14.0 Nobelium No 102 (253) Osmium Os 76 190.2 Oxygen O 8 16.0 Palladium Pd 46 106.4 Phosphorus P 15 31.0 Platinum Pt 78 195.1 Plutonium Pu 94 (242) Polonium Po 84 (209) Potassium K 19 39.1 Praseodymium Pr 59 140.9 Promethium Pm 61 (145) Protactinium Pa 91 (231) Radium Ra 88 (226) Radon Rn 86 (222) Rhenium Re 75 186.2 Rhodium Rh 45 102.9 Rubidium Rb 37 85.5 Ruthenium Ru 44 101.1 Samarium Sm 62 150.4 Scandium Sc 21 45.0 Selenium Se 34 79.0 Silicon Si 14 28.1 Silver Ag 47 107.9 Sodium Na 11 23.0 Strontium Sr 38 87.6 Sulfur S 16 32.1 Tantalum Ta 73 180.9 Technetium Tc 43 (98) Tellurium Te 52 127.6 Terbium Tb 65 158.9 Thallium Tl 81 204.4 Thorium Th 90 232.0 Thulium Tm 69 168.9 Tin Sn 50 118.7 Titanium Ti 22 47.9 Tungsten W 74 183.8 Uranium U 92 238.0 Vanadium V 23 50.9 Xenon Xe 54 131.3 Ytterbium Yb 70 173.0 Yttrium Y 39 88.9 Zinc Zn 30 65.4 Zirconium Zr 40 91.2