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About Formulas. Types of Bonding Ionic Occurs between metals & nonmetals Electrons are transferred Form crystal lattice NaCl CaBr 2 Covalent Occurs between.

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Presentation on theme: "About Formulas. Types of Bonding Ionic Occurs between metals & nonmetals Electrons are transferred Form crystal lattice NaCl CaBr 2 Covalent Occurs between."— Presentation transcript:

1 About Formulas

2 Types of Bonding Ionic Occurs between metals & nonmetals Electrons are transferred Form crystal lattice NaCl CaBr 2 Covalent Occurs between nonmetal Electrons are shared Form molecules H 2 O C 6 H 12 O 6 Metallic Occurs between metals Electrons are shared Form crystal lattice Al Fe

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

8 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

9 Formulas Ex: 2H 2 O can mean Ex: 2H 2 O can mean 2 molecules of water2 molecules of water 2 moles of water.2 moles of water. 2H 2 O molecules have 4 hydrogen atoms & 2 oxygen atoms. 2 Moles of H 2 O molecules have 4 moles of hydrogen atoms & 2 moles of oxygen atoms.

10 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.

11 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.

12 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.

13 Say everything you can about: NaClC 6 H 6 C 6 H 12 O 6 CH 4 CF 4 BeSO 4 C 8 H 18 KIC 2 H 4 Cl 2 CaBr 2 CuSO 4 5H 2 O Ionic, empirical Covalent, empir., molec? Covalent, molecular Ionic, empirical Covalent, molecular Covalent, empirical, molecular? Ionic, empirical Covalent, molecular Ionic, empirical Covalent, molecular

14 Say everything you can about PH 3 C 2 H 4 Al 2 O 3 SrI 2 NF 3 H 2 Se CH 3 OHSiO 2 H 2 O 2 CCl 4 XeF 4 P 4 O 10

15 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.

16 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

17 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.

18 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

19 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 KCrO 3.5 ? No dec’s Multiply ‘til whole: K 2 Cr 2 O 7. Multiply all three by 2, doesn’t change relationship.

20 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.

21 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.

22 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 = C 6 H 6. Empirical mass

23 What can you say about CuSO 4  5H 2 O? It’s a hydrated salt. For every mole of CuSO 4, 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, CuSO 4. CuSO 4 5H 2 O  CuSO 4 + 5H 2 O 

24 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 CuSO 4 5H 2 O, how much water would you lose on heating? If you had 5 moles of CuSO 4 5H 2 O, how much water would you lose? 

25 Percent Water in CuSO 4 5H 2 O Step 1: Calculate the formula mass. Mass of salt + Mass of water Percent = (Part/Whole) X 100% % H 2 O = Mass H 2 O/Formula Mass X 100 % % composition from the formula

26 Percent Water in CuSO 4 5H 2 O 12.00 grams of CuSO 4 5H 2 O are heated gently. The final mass after repeated heatings is 7.68 grams. What is the % of water? Mass of H 2 O = (12.00 – 7.68) = 4.32 g Percent H 2 O = 4.32 g X 100% = 36% 12.00 g Percent composition from experimental data

27 CuSO 4  xH 2 O  CuSO 4 + xH 2 O Use experimental data to find x: 12.00 grams of CuSO 4 5H 2 O are heated gently. (Hydrated salt.) 7.68 g = mass anhydrous salt remaining = mass CuSO 4. Mass of H 2 O = (12.00 – 7.68) = 4.32 g 

28 CuSO 4  xH 2 O  CuSO 4 + xH 2 O Note: moles of CuSO 4 & moles of H 2 O can be determined. Have grams already. Convert to moles Divide by small Multiply ‘til whole. 

29 CuSO 4  xH 2 O  CuSO 4 + xH 2 O 7.68 g CuSO 4 / 159.6 g/mol = 0.04812 mol CuSO 4 4.32 g H 2 O / 18.0 g/mol = 0.240 mol H 2 O Divide by small, which is 0.04812 CuSO 4 5H 2 O 


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