Writing Formulas Using the Crossover Method Examples

Writing Formulas Using the Crossover Method Examples
The crossover method is a shortcut way of writing formulas for ionic compounds if you know the charges on the ions in the compound. We’ll go over a few examples

a) Write the formula for magnesium chloride
Mg Cl– MgCl2 In example (a), we’re asked to write the formula for magnesium chloride.

Mg Cl– MgCl2 Looking up Magnesium on the ion table or periodic table, we see its Mg with charge of 2+.

Mg Cl– MgCl2 And chloride is Cl with a charge of minus, or minus 1.

Mg Cl– Mg1Cl2 We start the formula by writing down the symbols, Mg and Cl

Mg Cl– Mg1Cl2 Now we take the 2 from the charge on magnesium, (click) cross over, and (click) write it to next to the chloride.

Mg Cl– Mg1Cl2 Now we could take the 1 from the charge on chloride, (click) cross over, and write it next to the magnesium, but we don’t use the subscript 1 in a formula, so we don’t need to write anything.

Mg Cl– Mg1Cl2 So the formula has one Mg and two Cl’s.

Mg Cl– MgCl2 Formula So we’ll compact it and write the final formula for magnesium chloride as MgCl2.

Mg Cl– MgCl2 2 Formula We’ll check the charges to verify that this formula is correct. (click) We have one Mg2+ ion and 2 Cl– ions

(–1) Mg Cl– MgCl2 2 Formula So the total charge is positive 2, plus 2 × –1, or positive 2 plus negative 2,

(–1) = 0 Mg Cl– MgCl2 2 Formula Which is equal to zero, so the formula MgCl2 is correct.

b) Write the formula for calcium oxide
Ca O2– Ca2O2 Example b asks us to write the formula for calcium oxide.

Ca O2– Ca2O2 We look up calcium and its Ca with a charge of positive 2.

Ca O2– Ca2O2 And oxide is “O” with a charge of negative 2.

Ca O2– Ca2O2 We start the formula by writing down the symbols Ca and O.

Ca O2– Ca2O2 The charge on calcium is positive 2, so we (click) cross over and (click) write a 2 next to Oxygen in the formula

Ca O2– Ca2O2 The charge on oxide is negative 2, so we (click) cross over and (click) write a 2 next to Calcium in the formula. Notice we always drop the negative when we write subscripts

Ca O2– Ca2O2 Now, we have the formula Ca2O2.

Ca2+ O2– Ca2O2 b) Write the formula for calcium oxide
Both subscripts can be divided by 2 But you can see that both of the subscripts on this formula are divisible by 2.

b) Write the formula for calcium oxide Divide both subscripts by 2
Ca O2– Ca2O2 1 1 Divide both subscripts by 2 So we’ll simplify and divide both of these by 2 and get 1’s

Ca O2– Ca2O2 1 1 Remove the 1’s We do not keep the subscript 1 in a formula, so we (click) remove the 1’s, leaving us with the formula CaO.

Ca O2– CaO2 Formula Which we’ll compact and write as the final formula for calcium oxide.

Ca O2– CaO2 Formula Checking charges, we have one Ca2+ ion and one O2– ion.

–2 Ca O2– CaO2 Formula So the total charge is positive 2 plus negative 2,

–2 = 0 Ca O2– CaO2 Formula Which is equal to zero. So this formula is correct.

c) Write the formula for chromium(III) sulphide
Cr S2– Cr2S32 Question c asks us to write the formula for chromium(III) sulphide.

Cr S2– Cr2S32 The chromium (III) ion is Cr 3 plus

Cr S2– Cr2S32 And the sulphide ion is S 2 minus

Cr S2– Cr2S32 We start the formula with the symbols Cr and S

Cr S2– Cr2S32 We take the 3 from the charge on chromium, (click) cross over and write (click) 3 next to sulphur in the formula.

Cr S2– Cr2S32 And we take the 2 from the 2 minus charge on the sulphide ion and (click) cross over and write (click) 2 next to chromium in the formula.

Cr S2– Cr2S32 There is no number other than 1, that will divide into 2 and 3, so this formula cannot be simplified.

Cr S2– Cr2S32 Formula So the final formula for chromium III sulphide is Cr2S3.

2 Cr S2– Cr2S32 3 Formula Checking charges, we have 2 Cr3+ ions and 3 S2– ions,

2(+3) (–2) 2 Cr S2– Cr2S32 3 Formula So the total charge is 2 × positive 3, plus 3 × (–2), or positive 6 plus –6,

2(+3) (–2) = 0 2 Cr S2– Cr2S32 3 Formula Which is equal to zero, so this formula is correct.

d) Write the formula for palladium(IV) oxide
Pd O2– Pd2O42 Question d asks us to write the formula for palladium(IV) oxide.

Pd O2– Pd2O42 Palladium(IV) is Pd with a charge of 4 plus

Pd O2– Pd2O42 And oxide is “O” with a 2 minus charge

Pd O2– Pd2O42 We start the formula with the symbols Pd and O

Pd O2– Pd2O42 We take the 4 from the charge on palladium, (click) crossover and (click) write 4 next to oxygen in the formula

Pd O2– Pd2O42 We take the 2 from the charge on the oxide ion, (click) crossover and (click) write 2 next to palladium in the formula

Pd O2– Pd2O42 Now, we have the formula Pd2O4.

Pd4+ O2– Pd2O42 d) Write the formula for palladium(IV) oxide
Both subscripts can be divided by 2 But we see that both of the subscripts 2 and 4 are divisible by 2.

Pd4+ O2– Pd2O42 1 2 d) Write the formula for palladium(IV) oxide
Divide both subscripts by 2 So we simplify by dividing both the 2 and the 4 by 2, giving us 1 and 2.

Pd O2– Pd2O42 1 2 Remove the 1 We remove the subscript 1

Pd O2– Pd2O42 1 2 Leaving us with the formula PdO2.

Pd O2– PdO2 Formula Which we’ll compact and state as the final formula for palladium(IV) oxide

Pd O2– PdO2 2 Formula Checking charges, we have one Pd4+ ion and two O2– ions

(–2) Pd O2– PdO2 2 Formula So the total charge is positive 4 plus 2 times –2, or positive 4 plus negative 4,

(–2) = 0 Pd O2– PdO2 2 Formula Which is equal to zero, so PdO2 is the correct formula.

Writing Formulas Using the Crossover Method Examples

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Writing Formulas Using the Crossover Method Examples

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