Chemical analysis and the dilute hydrochloric acid group
الأسبوع 6 · الباب الثاني «التحليل الكيميائي» (١): مقدمة الباب ومراجعة قوانين المول (تدريب من إعداد Darslab) وأهمية التحليل الكيميائي والتحليل الوصفي والكمي ومجموعات الأنيونات الثلاث ومجموعة حمض الهيدروكلوريك المخفف (الكربونات والبيكربونات والكبريتيت والكبريتيد) · 3 دروس
1
Chemical analysis: what it is for, and a revision of mole formulas
Chemistry Sec 3 Term 1, Chapter 2: pp. 28–35
GoalSay what qualitative and quantitative analysis ask, and use the mole formulas on new numbers (Darslab practice).
New words
qualitative analysis · finds out which substances or ions are presentquantitative analysis · measures how much of a substance is present; it comes after the qualitative stepanion · the negative ion of a salt, such as Cl⁻ or CO₃²⁻
Explanation
Chemical analysis asks two questions. Qualitative analysis asks what is there. Quantitative analysis asks how much, and it comes second, because we cannot measure an amount before we know what to measure. A pure substance can be recognised by physical constants: its melting point, its boiling point or its molar mass, while a mixture is separated first. Organic analysis finds the elements and functional groups of a compound; inorganic analysis finds the ions of a salt, the anions (negative) and the cations (positive), and this course studies only the second kind. Laboratories use analysis in industry, farming, the environment and medicine, for example to measure substances in water, soil, blood or urine. Quantitative work needs the mole formulas, so this lesson revises them first with new numbers. This revision is Darslab practice, not a copy of the book. Keep units on every line.
Quantity
Formula
Unit
amount of substance
n = m ÷ M
mol
volume of a gas at STP
V = n × 22.4
L
density of a gas at STP
d = M ÷ 22.4
g/L
molarity
C = n ÷ V (V in litres)
mol/L
percentage
part ÷ whole × 100
%
moles of electrons (z = charge)
n(e⁻) = n(ion) × z
mol
Worked examples
ƒTwo steps
n = 5.3 ÷ 106 = 0.05 mol C = 0.05 ÷ 0.25 = 0.2 mol/L
Moles and molarity. A sample of 5.3 g of sodium carbonate, Na₂CO₃, is dissolved to make 250 mL of solution (Na = 23, C = 12, O = 16). M = 2 × 23 + 12 + 3 × 16 = 106 g/mol. n = 5.3 ÷ 106 = 0.05 mol. The volume is 0.250 L, so C = 0.05 ÷ 0.25 = 0.2 mol/L.
A gas at STP. Take 22 g of carbon dioxide, CO₂ (C = 12, O = 16, 22.4 L per mole at STP). M = 44 g/mol, so n = 22 ÷ 44 = 0.5 mol. V = 0.5 × 22.4 = 11.2 L. The density is d = 44 ÷ 22.4 = 1.96 g/L, which is more than air, so the gas sinks.
Percentages. In sodium chloride, NaCl (Na = 23, Cl = 35.5), M = 58.5 g/mol, so the mass percentage of sodium is 23 ÷ 58.5 × 100 = 39.3 %. In a different job, a 5.0 g sample of rock salt contains 4.2 g of NaCl: purity = 4.2 ÷ 5.0 × 100 = 84 %. The first percentage belongs to the compound; the second belongs to the sample.
Class activity
Kitchen scale analysis, with a grown-up. Weigh one level teaspoon of table salt on a kitchen scale and write the mass in grams. Use M = 58.5 g/mol to find the moles, then imagine the salt dissolved to make 250 mL (a cup) and work out the concentration. Do not taste the water, and say that your numbers are real measurements from your own scale, not book numbers. Then write two cards: one question that is qualitative (for example, is there salt in this water?) and one that is quantitative (how much salt is in each litre?).
Practice
1
A sample of 4.2 g of sodium hydrogencarbonate, NaHCO₃ (Na = 23, H = 1, C = 12, O = 16), is dissolved to make 500 mL. Find n and the concentration.
M = 84 g/mol, n = 4.2 ÷ 84 = 0.05 mol, C = 0.05 ÷ 0.5 = 0.1 mol/L.
2
For 8 g of oxygen gas, O₂ (O = 16, 22.4 L per mole at STP), find the volume at STP and the density.
n = 8 ÷ 32 = 0.25 mol, V = 0.25 × 22.4 = 5.6 L, d = 32 ÷ 22.4 = 1.43 g/L.
3
Find the mass percentage of calcium in calcium carbonate, CaCO₃ (Ca = 40, C = 12, O = 16). Then find the purity of a 2.5 g limestone sample that contains 2.0 g of CaCO₃.
