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Writing Chemical Formulas In this presentation you will: explore how to write chemical formulas Next >

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Presentation on theme: "Writing Chemical Formulas In this presentation you will: explore how to write chemical formulas Next >"— Presentation transcript:

1 Writing Chemical Formulas In this presentation you will: explore how to write chemical formulas Next >

2 Why is the formula for sodium chloride NaCl? Next > Why is it not Na 2 Cl 2 or Na 3 Cl? Introduction There are certain rules that determine the way in which elements can combine together. We will look at the rules, and then see how we can write the chemical formulas for different types of compounds.

3 Next > Oxidation Number The oxidation number is the charge that an atom would have, if the compound was composed of ions. The key to combining elements is the oxidation number (ON). It is also known as the oxidation state. ElementAnion fluorideF-F- chlorideCl - bromideBr - iodide I-I- oxideO 2- sulfideS 2- nitrideN 3- ElementCation lithiumLi + sodiumNa + potassiumK+K+ cesiumCs + magnesiumMg 2+ calciumCa 2+ aluminumAl 3+ copper(I)Cu + silverAg + copper(II)Cu 2+ iron(II)Fe 2+ lead(II)Pb 2+ zincZn 2+

4 Next > Oxidation Number A polyatomic ion, is an ion that contains more than one type of atom. An ion, is an atom that has lost or gained, one or more electrons. For example: an ion is Br -, a polyatomic ion is SO 4 2-. ElementAnion fluorideF-F- chlorideCl - bromideBr - iodide I-I- oxideO 2- sulfideS 2- nitrideN 3- ElementCation lithiumLi + sodiumNa + potassiumK+K+ cesiumCs + magnesiumMg 2+ calciumCa 2+ aluminumAl 3+ copper(I)Cu + silverAg + copper(II)Cu 2+ iron(II)Fe 2+ lead(II)Pb 2+ zincZn 2+

5 Oxidation Number Some elements can have more than one oxidation number. That means they can combine with the same element in more than one way. For example, copper can exist with one extra electron or two extra electrons. Cu + or Cu 2+. It can then combine with oxygen to produce two different chemicals: Cu 2 O or CuO. Next >

6 Oxidation Number Rule 2: For simple ions, the oxidation number is the charge on the ion. Rule 1: Atoms of a pure element have an oxidation number of 0. Rule 3: For compounds, the sum of the oxidation numbers is zero. The elements sodium, Na, carbon, C, and fluorine, F, have an ON of 0. The ions sodium, Na +, chlorine, Cl -, and magnesium, Mg 2+, have ONs of +1, and +2 respectively. In the compound KBr, the bromide ion has an ON of - 1. The potassium ion must therefore have a ON of +1. Example

7 Next > Oxidation Number Rule 4: For polyatomic ions, the sum of the oxidation numbers is the charge on the ion. In the ClO - ion, oxygen has an ON of -2. The chlorine ion must have an ON of +1 to give an overall ON of Example

8 Ion Oxidation number H +, Li +, Na +, K + +1 Mg 2+, Ca 2+, Ba 2+ +2 Al 3+ +3 O 2- -2 F-F- Cu( ɪ ) +1 Cu( ɪɪ ) +2 Next > Oxidation Number We can regard that as a fixed oxidation number. Some ions only have one oxidation number. Some ions have a variable oxidation number. These are represented by a Roman numeral shown in brackets immediately after the name. For example, copper(I) and copper(II).

9 Next > Finding the Formula Step 1: Write down the two parts with their oxidation numbers below them. What is the formula for sodium oxide compound? oxidation numbers

10 Next > Finding the Formula What is the formula for sodium oxide compound? Step 2: Since the mathematical sum of the oxidation numbers must be zero (Rule 3), multiply one or other term so that the terms are equal, but with opposite sign. This gives us the final formula Na 2 O.

11 Next > Finding the Formula Step 1: What is the formula for copper(II) oxide compound? Step 2: Not needed as they are already equal. This gives us the final formula CuO.

12 Next > Finding the Formula What is the formula for copper(I) oxide compound? Step 2: This gives us the final formula Cu 2 O. Step 1:

13 Question 1 What is the formula for lead(IV) oxide? Remember oxygen has an oxidation number of –2. A) PbO B) Pb 3 O 4 C) Pb 2 O 3 D) PbO 2 Next >

14 Question 1 What is the formula for lead(IV) oxide? Remember oxygen has an oxidation number of –2. A) PbO B) Pb 3 O 4 C) Pb 2 O 3 D) PbO 2

15 Next > Covalent Compounds In a covalent compound, the name gives us all the information we need. What is the formula for sulfur dioxide? This gives us the final formula SO 2. The di- prefix tells us that there are two of them. ElementAnion fluorideF-F- chlorideCl - bromideBr - iodide I-I- oxideO 2- sulfideS 2- nitrideN 3- ElementCation lithiumLi + sodiumNa + potassiumK+K+ cesiumCs + magnesiumMg 2+ calciumCa 2+ aluminumAl 3+ copper(I)Cu + silverAg + copper(II)Cu 2+ iron(II)Fe 2+ lead(II)Pb 2+ zincZn 2+

16 Question 2 What is the formula for dichloride monoxide? A) Cl 2 O B) ClO 2 C) ClO D) Cl 2 O 2 Next >

17 Question 2 What is the formula for dichloride monoxide? A) Cl 2 O B) ClO 2 C) ClO D) Cl 2 O 2

18 Next > Polyatomic Ions A polyatomic ion is one containing more than one element. What is the formula for the ammonium ion that has a charge of +1? Ammonia contains nitrogen and hydrogen: N and H. We know the hydrogen ion always carries a +1 oxidation number.

19 Next > Polyatomic Ions This gives the final formula NH 4 +. Nitrogen has numerous oxidation numbers; the most common ones are -3, +3 and +5. For it to combine with hydrogen it must be oppositely charged, so its oxidation number is -3. NH 3 would produce a neutral atom, but NH 4 would give a charge of +1.

20 Question 3 Given that the chromate ion has a charge of -2, determine its formula. The chromium ion has an oxidation number of +6, and oxygen has an oxidation number of -2. A) CrO -2 B) CrO 2 -2 C) CrO 4 -2 D) CrO 3 -2 Next >

21 Question 3 Given that the chromate ion has a charge of -2, determine its formula. The chromium ion has an oxidation number of +6, and oxygen has an oxidation number of -2. A) CrO -2 B) CrO 2 -2 C) CrO 4 -2 D) CrO 3 -2

22 Next > Common Acids Binary acids Binary acids are aqueous solutions of hydrogen and a non-metal. This gives us the final formula H 2 S. What is the formula for hydrosulfuric acid? This must be a hydrogen ion, H+,H+, together with a sulfide ion, S 2-. To make the resulting oxidation number zero, we need two hydrogen ions and one sulfide ion.

23 Next > Common Acids Oxyacids These are acids that have an ion containing oxygen in them. This gives us the final formula H 2 SO 4. What is the formula for sulfuric acid? Step 1: Change the –ic to the –ate. (If it was sulfurous acid we would change it to sulfite.) That means we have a hydrogen ion and a sulfate ion. That is H+ H+ and SO 4 2- Step 2: Equate the charges.

24 Question 4 Give the formula for phosphoric acid. Hydrogen has an oxidation number of +1 and the phosphate ion, PO 4 has a charge of -3. A) H 2 PO 4 B) H 3 PO 4 C) H 3 (PO 4 ) 2 D) H 4 PO 4 Next >

25 Question 4 Give the formula for phosphoric acid. Hydrogen has an oxidation number of +1 and the phosphate ion, PO 4 has a charge of -3. A) H 2 PO 4 B) H 3 PO 4 C) H 3 (PO 4 ) 2 D) H 4 PO 4

26 Next > Common Bases Bases are almost all hydroxides. What is the formula for ammonium hydroxide? This gives the final formula as NH 4 OH. The ammonium ion is written as NH 4 +. The hydroxide ion is OH -. Their oxidation numbers are +1 and respectively.

27 Question 5 Give the formula for copper(I) hydroxide. This combines Cu + with (OH) -. A) CuOH B) Cu 2 OH C) Cu(OH) 2 D) Cu 2 (OH) 2 Next >

28 Question 5 Give the formula for copper(I) hydroxide. This combines Cu + with (OH) -. A) CuOH B) Cu 2 OH C) Cu(OH) 2 D) Cu 2 (OH) 2

29 how to write formulas In this presentation you have seen: End > Summary


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