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#10 Write formulas for compounds formed from these pairs of ions: a. Ba 2+ and S -2

#10 Write formulas for compounds formed from these pairs of ions: a. Ba 2+ and S -2. First, crisscross the charges to balance the negatives vs positives. Ba +2 S - 2. 2. 2. Then, lower the numbers if possible to achieve the lowest whole number ratio. BaS. b. Li + and O -2.

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#10 Write formulas for compounds formed from these pairs of ions: a. Ba 2+ and S -2

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  1. #10 Write formulas for compounds formed from these pairs of ions: a. Ba 2+ and S-2 First, crisscross the charges to balance the negatives vs positives Ba+2 S-2 2 2 Then, lower the numbers if possible to achieve the lowest whole number ratio BaS b. Li+ and O-2 Li+ O-2 2 1 Li2O

  2. #10 Write formulas for compounds formed from these pairs of ions: c. Ca 2+ and N-3 First, crisscross the charges to balance the negatives vs positives Ca+2 N-3 3 2 Then, lower the numbers if possible to achieve the lowest whole number ratio Ca3N2 d. Cu+2 and I-1 Cu2+ I1- 1 2 CuI2

  3. #11 Write formulas for these compounds: a. sodium iodide b. stannous chloride stannous (unfortunately) is the old “classical” name for the lower of the two charges that tin can take. According to the chart on p. 255, Sn (tin) can be +2 or +4 stannous is the smaller number; +2. The other charge, +4 would be called stannic (no roman numbers are used in the “classical” system) Na+1 and I-1 1 1 NaI Therefore, stannous chloride is: Sn+2 and Cl-1 or SnCl2

  4. #11c. potassium sulfide please note that this is sulfide which is just the element sulfur. If it were sulfate it would be SO4-2 If it were sulfite it would be SO3-2 K+1 and S-2 K2S d. calcium iodide Ca+2 and I-1 CaI2

  5. #12 Write formulas for compounds formed from these pairs of ions. (NH4)2SO3 a. NH4+1 and SO3-2 Always use parenthesis when you have more than one polyatomic ion (positive or negative) to avoid interpreting it as 42. b. calcium ion and phosphate ion Ca+2 and PO4-3 Ca3(PO4)2

  6. #13 Write formulas for these compounds: a. lithium hydrogen sulfate note: you get the name for the polyatomic anions from the chart on p. 257 Li+1 and HSO4-1 LiHSO4 b. chromium (III) nitrite Cr+3 and NO2-1 Cr(NO2)3 Cr(NO2)3

  7. #14 Key concept: Describe how to determine the names of binary ionic compounds. binary ionic compounds are simply two elements, one a cation and the other an anion coming together to make a compound examples: NaCl LiBr MgO etc. The rules are: 1. put the cation first and the anion last. 2. add an –ide ending to the anion. 3. if there is more than one charge possible on the cation, you must include a roman numeral in the name to show this: example: iron (II) oxide is FeO iron (III) oxide is Fe2O3

  8. #15 Key Concept: Describe how to write the formulas for binary ionic compounds. Rules for writing formulas: 1. write down the correct elements 2. determine the correct charges on those elements. 3. write a balanced compound with lowest whole number ratios. If you only have one charge possibility such as zinc (check the periodic table if you are in doubt) You do not need a roman numeral! example: zinc chloride write down Zn and Cl check their Zn+2 (only!) and Cl-1 charges write the balanced compound: ZnCl2

  9. #16 Key concept: How do you write the formulas and the names of compounds with polyatomic ions? Same as binary except you just put in the correct name of the polyatomic ion. look this anion up on the chart to get its proper name example: Na+1 and HCO3-1 is called sodium hydrogen carbonate NaHCO3

  10. #17 Write the formulas for these binary compounds: a. beryllium chloride b. cesium sulfide c. sodium iodide d. strontium oxide BeCl2 Cs2S NaI SrO

  11. #18 Write the formulas for these compounds containing polyatomic ions. a. chromium (III) nitrite b. sodium perchlorate c. magnesium hydrogen carbonate d. calcium acetate Cr(NO2)3 NaClO4 Mg(HCO3)2 Ca(C2H3O2)2

  12. #19 Identify any incorrect formulas. Explain your answer incorrect! Mg is +2 and SO4 is -2 so the correct formula should have been: a. Mg2(SO4)3 MgSO4 correct! Rb is +1 and As is -3 b. Rb3As incorrect! Be is +2 (only!) and Cl is -1 so the correct formula should have been: c. BeCl3 BeCl2 d. NaF correct! Na is +1 and F is -1

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