Percent comp and atomic mass(1)

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Atomic mass, percent composition, and practice problems
Practice You have 200.4 g of C 6 H 12 O 6 and want it to react fully with O 2 . Please calculate how many grams of O 2 you need. Also please let me know how many molecules of each product will be made (theoretical yield). O 2 + C 6 H 12 O 4 H 2 O + CO 2
Practice You have 200.4 g of C 6 H 12 O 6 and want it to react fully with O 2 . Please calculate how many grams of O 2 you need . Also please let me know how many molecules of each product will be made (theoretical yield). O 2 + C 6 H 12 O 4 H 2 O + CO 2
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Practice O 2 + C 6 H 12 O 6 H 2 O + CO 2 6O 2 + C 6 H 12 O 6 6H 2 O + 6CO 2 6(12.0107g/mol) + 12(1.00794g/mol) + 6(15.9994g/mol) = 180.15588 g/mol
Mass A Moles A Moles B Mass B 1 Mole A. molar mass A (g) Coeff. B Coeff. A molar mass B (g) 1 Mole B Molecules B 6.02 · 10 23 Molecules 1 Mole B Practice
Glucose 180.15588 g/mol 200.4g C 6 H 12 O 6 · · · = 213.6g O 2 200.4g C 6 H 12 O 6 · · · = 4.02·10 24 molecules of CO 2 200.4g C 6 H 12 O 6 · · · = 4.02·10 24 molecules of H 2 O
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Percent Composition
Atomic Mass Atomic mass is based on a relative scale and the mass of  12 C (carbon twelve) is defined as 12 amu. Why do we specify  12 C? We do not simply state the the mass of a C atom is 12 amu because elements exist as a variety of isotopes. Carbon exists as two major isotopes, 12 C, and 13 C ( 14 C exists and has a half life of 5730 y, 10 C and 11 C also exist; their half lives are 19.45 min and 20.3 days respectively). Each carbon atom has the same number of protons and electrons, 6. 12 C has 6 neutrons, 13 C has 7 neutrons, and 14 C has 8 neutrons and so on. Since there are a variety of carbon isotopes we must specify which C atom defines the scale. All the masses of the elements are determined relative to  12 C.
Isotope Atomic Mass amu Natural Abundance % 16 O 15.99491 99.759 17 O 16.99913 0.037 18 O 17.99916 0.204 If we know the natural abundance (the natural abundance of an isotope of an element is the percent of that isotope as it occurs in a sample on earth) of all the isotopes and the mass of all the isotopes we can find the average atomic mass. The average atomic mass is simply a weighted average of the masses of all the isotopes.  
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Percent Composition of Compounds Percent composition by mass is the percent by mass of each element in a compound. To calculate mass percent of an element, assume 1 mol of compound. mass of element mass due to that element total mass of compound % 100%
O 2 + C 6 H 12 O 6 H 2 O + CO 2 6O 2 + C 6 H 12 O 6 6H 2 O + 6CO 2 6(12.0107g/mol) + 12(1.00794g/mol) + 6(15.9994g/mol) = 180.15588 g/mol
6(12.0107g/mol) = 72.0642 72.0642 / 180.15588 g/mol = 0.400001 12(1.00794g/mol) = 12.09528 12.9528/ 180.15588 g/mol = 0.0718977 6(15.9994g/mol) / 180.15588 g/mol = 95.9964 / 180.15588 g/mol = 0.532852
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6(12.0107g/mol) = 72.0642 72.0642 / 180.15588 g/mol = 0.400001 = 40.0001% 12(1.00794g/mol) = 12.09528 12.9528/ 180.15588 g/mol = 0.0718977 = 7.18977% 6(15.9994g/mol) / 180.15588 g/mol = 95.9964 / 180.15588 g/mol = 0.532852 = 53.2852%
Calculate mass% of each element in K 2 CO 3 Use: K = 39 amu C = 12 amu O= 16 amu mass of element mass due to that element total mass of compound % 100%
Calculate mass% of each element in K 2 CO 3 Use: K = 39 amu 57 % C = 12 amu 8.7 % O= 16 amu 35 %
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Average Atomic Mass
Isotope Symbols For a given isotope, Z = atomic # = # protons A = mass # = # protons + # neutrons # neutrons = A – Z # electrons = # protons ex) , , or Magnesium-25 Z =12 = 12 protons A = 25 = protons + neutrons # neutrons = 25 – 12 = 13 neutrons # electrons = # protons = 12 12 Mg 24.305
12 Mg 24.305 Isotopes of Magnesium In naturally occurring Mg, there are 3 isotopes. Isotopes: Same number protons, DIFFERENT number neutrons Atomic mass (periodic table) is the weighted average of the isotope masses Weighted Average = % abundance1 * mass 1 + % abundance 2 * mass 2 …etc.
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Atomic mass (periodic table) is the weighted average of the isotope masses Weighted Average = % abundance1 * mass 1 + % abundance 2 * mass 2 …etc. Weighted Average = 0.7870 * 23.99 + 0.1013 * 24..99 + 0.1117 * 25.98 24.3134583 = 0.7870 * 23.99 + 0.1013 * 24..99 + 0.1117 * 25.98 24.31 amu
63.9291422 * 0.4917 + 65.9260334 * 0.2773 + 66.9271273 * 0.0404 + 67.9248442 * 0.1845 + 69.9253193 * 0.0061 = AAM 31.43 39592 + 18.28 12891 + 2.70 385594292 + 12.53 21337549 + 0. 42 654444773 = 65.38 amu
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Practice Problems
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What is the chemical symbol for silver ? a)S b)Si c)Sr d)Ag e)Au
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What is the chemical symbol for silver ? a)S b)Si c)Sr d)Ag e)Au
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What is the chemical symbol for gold ? a)S b)Si c)Sr d)Ag e)Au
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What is the chemical symbol for gold ? a)S b)Si c)Sr d)Ag e)Au
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Determine the number of protons, neutrons, and electrons in each of the following. A. 13 C B. 48 Ti 3+ C. 8 Li
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Determine the number of protons, neutrons, and electrons in each of the following. A. 13 C 6 7 6 B. 48 Ti 3+ C. 8 Li
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Determine the number of protons, neutrons, and electrons in each of the following. A. 13 C 6 7 6 B. 48 Ti 3+ 222619 C. 8 Li
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Determine the number of protons, neutrons, and electrons in each of the following. A. 13 C 6 7 6 B. 48 Ti 3+ 222619 C. 8 Li 3 5 3
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What is the cation in FeCl 2 ? a. Iron ion b. Iron (II) ion c. Iron (III) ion
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What is the cation in FeCl 2 ? a. Iron ion b. Iron (II) ion c. Iron (III) ion
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What is the cation in Al 2 O 3 ? a. aluminum (III) ion b. aluminum (II) ion c. aluminum ion
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What is the cation in Al 2 O 3 ? a. aluminum (III) ion b. aluminum (II) ion c. aluminum ion
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The presence of sulfur in coal gives rise to the acid rain phenomenon. How many atoms are in 60.5 g of S? N A = 6.022 x 10 23 a. 3.13 X 10 -24 atoms b. 1.17 X 10 27 atoms c. 1.14 X 10 24 atoms d. 3.22 X 10 -21 atoms
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The presence of sulfur in coal gives rise to the acid rain phenomenon. How many atoms are in 60.5 g of S? N A = 6.022 x 10 23 a. 3.13 X 10 -24 atoms b. 1.17 X 10 27 atoms c. 1.14 X 10 24 atoms d. 3.22 X 10 -21 atoms
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How many Cl+ ions are present in 1.38 g MgCl 2 ?
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How many Cl+ ions are present in 1.38 g MgCl 2 ? 0
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How many Cl- ions are present in 1.38 g MgCl 2 ? 24.305 + (2* 35.45) = 24.305 + 70.90 = 95.20 5 amu 1.38 g/ 95.20 5 g/mol = 0.0 144 9504 moles 0.0 144 9504 * 2 * 6.022 10 23 = 1.75 10 22 Cl- ions
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A major textile dye manufacturer developed a new yellow dye. The dye has a percent composition of 75.95% C, 17.72% N, and 6.33% H by mass with a molar mass of about 240 g/mol. Determine the molecular formula of the dye.
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A major textile dye manufacturer developed a new yellow dye. The dye has a percent composition of 75.95% C, 17.72% N, and 6.33% H by mass with a molar mass of about 240 g/mol. Determine the molecular formula of the dye. 15 carbon 3 nitrogen 18 hydrogen For carbon 240 g/mol * 0.7595 = 15.19 Can’t have 0.19 of a carbon so we round to 15
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