Find the by mass of oxygen in water

Find the % by mass of oxygen in water • Two things you need to know H 2 O – Mass of part – Mass of whole 2 atoms of hydrogen 2 x 1. 0 g/mol 2. 0 g/mol hydrogen 1 atom of oxygen 1 x 16. 0 g/mol oxygen 2. 0 g/mol + 16. 0 g/mol = 18. 0 g/mol for H 2 O 16. 0 g/mol 18. 0 g/mol x 100% = 88. 9% oxygen Episode 702

Lab Excerise • We are going to react sodium bicarbonate with hydrochloric acid Na. HCO 3 + HCl Na. Cl + CO 2 + H 2 O 1. Measure and determine the mass of CO 2 produced in the reaction 2. Determine the % carbon in the CO 2 produced 3. Determine the % carbon in the Na. HCO 3 used Episode 702

Find the % carbon in sodium bicarbonate (Na. HCO 3) Remember: (mass of element in 1 mol of compound ÷ molar mass of compound) x 100% Na H C O 1 x 23 g/mol = 23 g/mol 1 x 1 g/mol = 1 g/mol 1 x 12 g/mol = 12 g/mol 3 x 16 g/mol = 48 g/mol 84 g/mol Na. HCO 3 12 g/mol 84 g/mol x 100% = 14. 3% carbon Episode 702

Find the % composition of aluminum oxide • Write formula Al 2 O 3 • Calculate molar mass Al O 2 x 27 g/mol = 3 x 16 g/mol = 54 g/mol 48 g/mol 102 g/mol 54 g/mol x 100% = 102 g/mol 52. 9% aluminum 48 g/mol x 100% = 102 g/mol 47. 1% oxygen Episode 702

Empirical Formula • Simplest whole number ratio of elements in a compound – Example • Find the empirical formula for a compound containing 56. 6 g K, 8. 7 g C, and 34. 7 g O Step 1: Convert each mass into moles of the element. 56. 6 g K 1 mol K = 1. 45 mol K 39. 1 g K 8. 7 g C 1 mol C = 0. 73 mol C 12 g C 34. 7 g. O 1 mol O = 2. 17 mol O 16 g. O K 1. 45 C 0. 73 O 2. 17 Step 2: Divide by the smallest to find a simple whole number. K 2 CO 3 Episode 702

32. 37% Na 22. 64% S 44. 99% O 100 % 32. 37 g Na 22. 64 g S 44. 99 g O 100 g 1. 41 mol Na 0. 705 mol S 2. 81 mol O (Hint: When % are given, assume you have 100 g of the compound, and the % changes to grams. ) Step 1: Convert each mass into moles of the element. Na 1. 41 S 0. 705 O 2. 81 Step 2: Divide by the smallest to find a simple whole number. Na 2 SO 4 Episode 702

Px. Oy Determine the grams of oxygen in the sample. 10. 15 g sample 10. 15 g – 4. 433 g = 5. 717 g oxygen 4. 433 g P Step 1: Convert each mass into moles of the element. 4. 433 g P 1 mol P = 0. 143 mol P 31 g P 5. 717 g O 1 mol O = 0. 357 mol O 16 g O P 0. 143 O 0. 357 Step 2: Divide by the smallest to find a simple whole number. PO 2. 5 (Hint: If the ratio is still not whole numbers, multiply both subscripts by a number, such as “ 2” to get rid of fraction. ) P 2 O 5 Episode 702