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Understanding Chemical Formulas: Metals vs. Nonmetals, Ionic vs. Covalent Bonding & Empirical to Molecular Conversion

Explore the differences between metals and nonmetals, crystal lattice versus molecules, atomic ratios in compounds, ionic and covalent formulas, and conversion of empirical to molecular formulas. Learn how to determine empirical formulas from percentage compositions and how to find molecular formulas from empirical formulas. Discover the concept of hydrated salts and predicting water loss upon heating. Dive into the relationship between empirical and molecular formulas and understand the calculations involved.

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Understanding Chemical Formulas: Metals vs. Nonmetals, Ionic vs. Covalent Bonding & Empirical to Molecular Conversion

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  1. About Formulas

  2. Types of Bonding

  3. Metals vs. Nonmetals • Most of the elements in the periodic table are metals. • Except for H, everything to the left of the staircase is a metal • Nonmetals are located to the right of the staircase.

  4. Crystal Lattice vs. Molecule • Molecule: Discrete unit, definite number of atoms. All molecules of a given compound are identical. Exact formula. • Crystal Lattice: Regular, repetitive, 3-D arrangement of atoms, ions, or molecules. No set size. Not perfect. No two exactly alike. • Formula gives smallest whole number ratios. (Formula unit). “Exact” formulas NOT useful.

  5. 2-D repetitive patterns

  6. NaCl Crystals: 3-D repetitive patterns

  7. Formulas • Tell the type & number of atoms in a compound. • Microscopic level: imagine 1 atom, molecule, or formula unit. • Formula gives atom ratios. • Macroscopic level: imagine working in the lab • Formula gives mole ratio

  8. Formulas • Ex: 2H2O can mean • 2 molecules of water • 2 moles of water. 2H2O molecules have 4 hydrogen atoms & 2 oxygen atoms. 2 Moles of H2O molecules have 4 moles of hydrogen atoms & 2 moles of oxygen atoms.

  9. Ionic vs. Covalent Formulas • By looking at the type of atoms, we can decide if the substance is an ionic compound or a molecular (covalent) compound. Molecular compounds usually have all nonmetals. Ionic compounds usually have metal + nonmetal.

  10. Empirical Formula • Smallest whole number ratio of the elements in the compound. • Ionic compounds have only empirical formulas. • Covalent compounds have empirical and molecular formulas. They can be the same or different.

  11. Molecular Formulas • For covalent (molecular) compounds. • Give the exact composition of the molecule. • Molecular compounds have both empirical and molecular formulas. They can be the same or different.

  12. Say everything you can about: NaCl C6H6 C6H12O6 CH4 CF4 BeSO4 C8H18 KI C2H4Cl2 CaBr2 CuSO45H2O Covalent, molecular Ionic, empirical Covalent, molecular Covalent, empir., molec? Covalent, empirical, molecular? Ionic, empirical Ionic, empirical Covalent, molecular Covalent, molecular Ionic, empirical Ionic, empirical

  13. Say everything you can about PH3 C2H4 Al2O3 SrI2 NF3 H2Se CH3OH SiO2 H2O2 CCl4 XeF4 P4O10

  14. Empirical Formulas from % Composition Convert to mass. Convert to moles. Divide by small. Multiply ‘til whole. Note 1: last step not always used. Note 2: sometimes you start at step 2, if they give analysis data in grams.

  15. Compound: 63.52% Fe & 36.48% S • Convert to grams: Assume 100 g sample-size to make math easy. • 63.52 g Fe and 36.48 g S • Convert to moles: Divide by atomic masses. 63.52 g Fe / 55.8 g Fe/mol Fe = 1.138 mol Fe 36.48 g S / 32.1 g S/mol S = 1.136 mol S

  16. Compound: 63.52% Fe & 36.48% S • Divide by small: This is where you take the ratio. • 1.136 and 1.138 are essentially the same. Divide both by 1.136 and get 1 for each. • Empirical formula is FeS.

  17. 26.56 % K, 35.41% Cr, & remainder O • Assume 100-g sample size. • 26.56 g K • 35.41 g Cr • 38.03 g O • Convert to moles. 26.56 g K / 39.1 g K/mol K = 0.679 mol K 35.41 g Cr / 52.0 g Cr/mol Cr = 0.681 mol Cr 38.03 g O / 16.0 g O/mol O = 2.38 mol O

  18. 26.56 % K, 35.41% Cr, & remainder O • Divide by small: • 0.679 and 0.681 are essentially the same. • Much smaller than 2.38. • Divide each by 0.68 • K: 1, Cr: 1, O: 3.5 KCrO3.5? No dec’s • Multiply ‘til whole: • K2Cr2O7. Multiply all three by 2, doesn’t change relationship.

  19. Relationship between empirical and molecular formulas • The molecular formula is a whole number multiple of the empirical formula. • Molec. Formula = n (Empirical Form.) • n is a small whole number, which multiplies the subscripts. Sometimes, n = 1.

  20. Molecular Formula • If you know the empirical formula and the molar mass, you can find the molecular formula. • Step 1: Find the mass of the empirical formula. • Step 2: Molar mass  Empirical mass = small whole number, n • Step 3: Multiply the subscripts in the empirical formula by n.

  21. Finding Molecular Formulas • Find the molecular formula for the substance whose empirical formula is CH and whose molar mass = 78.0 g • Step 1: Empirical mass = 13.0 g • Step 2: Molar mass = n = 78.0/13.0 • n = 6. • Step 3: 6 X subscripts in CH = C6H6. Empirical mass

  22. What can you say about CuSO45H2O? • It’s a hydrated salt. • For every mole of CuSO4, there are 5 moles of water. • If it’s heated, it dries out. The water goes into the air and you’re left with the anhydrous salt, CuSO4. CuSO45H2O  CuSO4 + 5H2O 

  23. CuSO45H2O  CuSO4 + 5H2O  • The mole ratio in the formula can be used to predict how much water would be given off by any size sample. • If you had 2 moles of CuSO45H2O, how much water would you lose on heating? • If you had 5 moles of CuSO45H2O, how much water would you lose?

  24. Percent Water in CuSO4•5H2O • Step 1: Calculate the formula mass. • Mass of salt + Mass of water • Percent = (Part/Whole) X 100% • % H2O = Mass H2O/Formula Mass X 100 % % composition from the formula

  25. Percent Water in CuSO4•5H2O 12.00 g 12.00 grams of CuSO4•5H2O are heated gently. The final mass after repeated heatings is 7.68 grams. What is the % of water? Mass of H2O = (12.00 – 7.68) = 4.32 g Percent H2O = 4.32 g X 100% = 36% Percent composition from experimental data

  26. CuSO4xH2O  CuSO4 + xH2O  Use experimental data to find x: 12.00 grams of CuSO4•5H2O are heated gently. (Hydrated salt.) 7.68 g = mass anhydrous salt remaining = mass CuSO4. Mass of H2O = (12.00 – 7.68) = 4.32 g

  27. CuSO4xH2O  CuSO4 + xH2O  • Note: moles of CuSO4 & moles of H2O can be determined. • Have grams already. • Convert to moles • Divide by small • Multiply ‘til whole.

  28. CuSO4xH2O  CuSO4 + xH2O  7.68 g CuSO4 / 159.6 g/mol = 0.04812 mol CuSO4 4.32 g H2O / 18.0 g/mol = 0.240 mol H2O Divide by small, which is 0.04812 CuSO45H2O

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