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Chemical Composition Chapter 6

Chemical Composition Chapter 6. Tro, 2 nd ed. DEFINITIONS OF VARIOUS MASSES. Formula or molecular mass = S of atomic masses in the chemical formula Molecular mass = mass in amu for a molecule, from nonmetal elements forming covalent bonds

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Chemical Composition Chapter 6

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  1. Chemical CompositionChapter 6 Tro, 2nd ed.

  2. DEFINITIONS OF VARIOUS MASSES Formula or molecular mass=S of atomic masses in the chemical formula Molecular mass= mass in amu for a molecule, from nonmetal elements forming covalent bonds Molecule is a covalent compound’s smallest unit, made of nonmetals in covalent bonds Formula mass= mass in amu for a formula unit or for all compounds Formula unit is the smallest whole number ratio of an ionic compound with metallic and nonmetallic elements in it (usually)

  3. Molecular and formula mass NaNO3 (NH4)2SO4 1 Na * 22.99 = 22.99 2 N * 14.01 = 28.02 1 N * 14.01 = 14.01 8 H * 1.008 = 8.064 3 O * 16.00 = 48.00 1 S * 32.07 = 32.07 85.00 amu 4 O * 16.00 = 64.00 132.15 amu

  4. Molecular and formula mass YOU PRACTICE: PCl5 HNO3 CHLORINE GAS

  5. ATOMS, MOLES & AVOGADRO’S NUMBER The mass of a single atom is too small to measure on a balance. mass of hydrogen atom = 1.673 x 10-24 g Chemists require a unit for counting which can express large numbers of atoms using simple numbers. Chemists have chosen a unit for counting atoms. That unit is the MOLE

  6. ATOMS, MOLES & AVOGADRO’S NUMBER 1 mole = 6.0221 x 1023 objects 6.0221 x 1023 objects/mole is Avogadro’s Number 6.0221 x 1023 is a very number LARGE

  7. ATOMS, MOLES & AVOGADRO’S NUMBER 1 mole of any element contains 6.0221 x 1023 particles of that substance. The atomic mass in grams of any element contains 1 mole of atoms. This is the same number of particles as there are in exactly 12 grams ofcarbon-12. 1 mole of atoms = 6.0221 x 1023 atoms 1 mole of molecules =6.0221 x 1023 molecules 1 mole of ions = 6.0221 x 1023ions 1 mole of formula units =6.0221 x 1023 formula units

  8. AVOGADRO’S NUMBER & MOLAR MASS The molar mass of an element is its atomic mass in grams INSTEAD of amu. It contains 6.0221 x 1023 atoms (Avogadro’s number) of the element. The units are grams/mole.

  9. Relating atomic mass and molar mass & Avogadro’s Number: This is per atom. This is per mole. This is per mole.

  10. THE MOLE MAP ATOMS, MOLECULES, FORMULA UNITS, IONS XNA /NA MOLES X M /M MASS (grams) / D X D VOLUME, mL

  11. ATOMS, MOLECULES& AVOGADRO’S NUMBER Practice: 1. How many atoms in 0.100 moles of He? 2. How many moles are represented by 9.00 x 109 atoms of He? 1. 0.100 mol * 6.0221 x 1023 atoms/mol = 6.0221 x 1022 He atoms 2. 9.00x109atoms = 1.50 x 10-14 mol of He 6.0221 x 1023 atoms/mol OR 9.00x109atoms * 1 mole = 1.50 x 10-14 mol He 6.0221 x 1023 atoms

  12. MOLAR MASS – JUST LIKE CALCULATING MOLECULAR MASS 1. Find molar mass of liquid bromine, Br2 2. Find molar mass of ethanol CH3CH2OH (organic structural formula) 3. Find molar mass of ammonium phosphate. 1. 2 * 79.90 g/mol = 159.80 g/mol 2. 2 C * 12.01 = 24.02 g/mol 6 H * 1.008 = 6.048 g/mol 1 O * 16.00 = 16.00 g/mol 46.068 = 46.07 g/mol (sig figs) 3. 149.12 g/mol

  13. MASS, MOLECULES & THE MOLE: USING THE MOLE MAP EXAMPLE: Use mole map to find number of molecules in 0.4607 grams of ethanol. First use molar mass to go from mass to moles: 0.4607 * 1 mol = 0.01000 MOLES 46.07 g Then use moles and Avogadro’s Number to find number of molecules: 0.1000 mol * 6.0221 x 1023 molecules/mol = 6.022 x 1021 molecules

  14. MASS, MOLECULES & THE MOLE: USING THE MOLE MAP EXAMPLE: Use mole map to find number of grams of lead in 9.00 x 1018 atoms of lead. First use Avogadro’s Number to find moles: 9.OO X 1018 atoms * 1 mol = 1.494 X 10-5 mol 6.0221 * 1023 atoms Then use molar mass to find grams: 1.494 X 10-5 mol Pb * 207.2 g/mol = 3.10 X 10-3 g (back to 3 sig figs)

  15. COMPOUNDS, MOLAR MASS& AVOGADRO’S NUMBER Chemical formulas are written to tell us: - the number of atoms of each element - the type of compound (ionic, covalent, acid, or organic) - the structure of the compound

  16. COMPOUNDS, MOLAR MASS& AVOGADRO’S NUMBER NaNO3 (NH4)2SO4 PCl5 HNO3 1 Na atom 2 N atoms 1 P atom 1 H atom 1 N atom 8 H atoms 5 Cl atoms 1 N atom 3 O atoms 1 S atom 3 O atoms 4 O atoms ionic ionic covalent acid formula formula molecule molecule unit unit (until aq)

  17. These relationships are present when hydrogen combines with chlorine.

  18. How many oxygen atoms are present in 2.00 mol of oxygen molecules? Two conversion factors are needed: Conversion sequence: moles O2→ molecules O2 → atoms O

  19. How many nitrogen atoms are in 56.04 g of N2? The molar mass of N2 is 28.02 g/mol. Conversion sequence: g N2 → moles N2→ molecules N2 → atoms N

  20. ELEMENTAL COMPOSITION Elemental composition is the percent composition of a compound, which is the mass percent of each element in the compound Example: H2O is 11.19% H by mass & 88.81% O by mass percent = part * 100% whole Elemental composition of sodium chloride, NaCl: 1 Na*22.99= 22.99 g/mol (22.99/58.44)*100 = 39.34% 1 Cl*35.45 = 35.45 g/mol (35.45/58.44)*100 = 60.66% 58.44 g/mol SUM = 100.00%

  21. Percent Composition From Formula If the formula of a compound is known, a two-step process is needed to calculate the percent composition. Step 1Calculate the molar mass of the formula. Step 2Divide the total mass of each element in the formula by the molar mass and multiply by 100.

  22. Percent Composition From Formula You practice with ethanol: CH3CH2OH 2 C*12.01 = 24.02 g/mol (24.02/46.07)*100 = 52.14% 6 H*1.008 = 6.048 g/mol (6.048/46.07)*100 = 13.13% 1 O*16.00 = 16.00 g/mol (16.00/46.07)*100 = 34.73% = 46.07 g/mol SUM = 100.00%

  23. Percent Composition From Experimental Data Percent composition can be calculated from experimental data without knowing the composition of the compound. Step 1Calculate the mass of the compound formed. Step 2Divide the mass of each element by the total mass of the compound and multiply by 100.

  24. A compound containing nitrogen and oxygen is found to contain 1.52 g of nitrogen and 3.47 g of oxygen. Determine its percent composition. Step 1Calculate the total mass of the compound 1.52 g N 3.47 g O 4.99 g = total mass of compound

  25. Continued: N30.5% O69.5% Step 2Divide the mass of each element by the total mass of the compound formed.

  26. Mole Relationships inChemical Formulas Since we count atoms and molecules in mole units, we can find the number of moles of a constituent element if we know the number of moles of the compound

  27. Mole Relationships inChemical Formulas Class work: Skillbuilder 6.6 How many moles of oxygen are in 1.4 moles of sulfuric acid? How many grams of oxygen is this? 1.4 mol H2SO4 * 4 mol O/mol H2SO4 = 5.6 mol of oxygen 5.6 mol O * 16.00 g/mol = 89.6 grams of O

  28. Empirical Formula versus Molecular Formula The empirical formula or simplest formula gives the smallest whole-number ratio of the atoms present in a compound. The molecularformula is the true formula of a compound. The empirical formula gives the relative number of atoms of each element present in the compound. The molecular formula represents the total number of atoms of each element present in one molecule of a compound.

  29. Notice that these are all covalent compounds or diatomic elements. That’s because ionic formulas are by definition the lowest whole number ratio, with only a few exceptions like mercury (I) chloride, Hg2Cl2.

  30. CalculatingEmpirical Formulas Step 1 Assume a definite starting quantity (usually 100.0 g) of the compound, if the actual amount is not given, and express the mass of each element in grams. Step 2 Convert the grams of each element into moles of each element using each element’s molar mass. Step 3Divide the moles of atoms of each element by the moles of atoms of the element that had the smallest value. – If the numbers obtained are whole numbers, use them as subscripts and write the empirical formula. – If the numbers obtained are not whole numbers, go on to step 4.

  31. CalculatingEmpirical Formulas Step 4Multiply the values obtained in step 3 by the smallest numbers that will convert them to whole numbers Use these whole numbers as the subscripts in the empirical formula. The results of calculations may differ from a whole number. If they differ ±0.05, round off to the next nearest whole number. Deviations greater than 0.05 unit from a whole number usually mean that the calculated ratios have to be multiplied by a whole number.

  32. Resulting WholeNumber 1 1 2 1 3 Some Common Fractions and Their Decimal Equivalents DecimalEquivalent Common Fraction 0.25 0.333… Multiply the decimal equivalent by the number in the denominator of the fraction to get a whole number. 0.666… 0.5 0.75

  33. The analysis of a salt shows that it contains 56.58% potassium (K); 8.68% carbon (C); and 34.73% oxygen (O). Calculate the empirical formula for this substance. Step 1 Express each element in grams. Assume 100 grams of compound. K = 56.58 g C = 8.68 g O = 34.73 g

  34. Continued: C has the smallest number of moles Step 2 Convert the grams of each element to moles.

  35. Continued: Step 3 Divide each number of moles by the smallest value. The simplest ratio of K:C:O is 2:1:3 Empirical formula K2CO3

  36. Now you try: The percent composition of a compound is 25.94% nitrogen (N), and 74.06% oxygen (O). Calculate the empirical formula for this substance. N2O5

  37. Calculating the Molecular Formula from the Empirical Formula The molecular formula can be calculated from the empirical formula if the molar mass is known. The molecular formula will be equal to the empirical formula or some multiple, n, of it. To determine the molecular formula evaluate n. n is the number of units of the empirical formula contained in the molecular formula.

  38. What is the molecular formula of a compound which has an empirical formula of CH2 and a molar mass of 126.2 g? Calculate the mass of each CH2 “mol” 1 C = 1(12.01 g/mol) = 12.01g/mol 2 H = 2(1.01 g/mol) = 2.02g/mol 14.03g/mol The molecular formula is (CH2)9 = C9H18

  39. Group Work: What is the empirical formula and the molecular formula of lactic acid, which is 40.00% C, 6.71% H and 53.29% O, with a molar mass of 90.1 g? C3H6O3 (or CH3CHOHCOOH)

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