Structure Bonding And Introduction To Organic Chemistry All subtopics

empirical formula is the simplest whole-number ratio of atoms of each element in a compound; molecular formula is the actual number of atoms of each element in one molecule

C₂H₅

a molecule is two or more atoms covalently bonded together; a compound contains two or more different elements chemically bonded in fixed proportions

6.02 × 10²³ mol⁻¹

Finding moles of carbon $$n=\frac{m}{{M}_{r}}=\frac{6.00}{12.0}=0.500\text{ mol}$$ Finding the number of atoms $$N=nL=0.500\times 6.02\times {10}^{23}$$ $$N=3.01\times {10}^{23}\text{ atoms}$$

Each CO₂ contains 2 O atoms, so the moles of oxygen are $${n}_{\text{O}}=0.250\times 2=0.500\text{ mol}$$ Finding the number of atoms $$N=0.500\times 6.02\times {10}^{23}$$ $$N=3.01\times {10}^{23}\text{ atoms}$$

C₃H₈ + 5O₂ → 3CO₂ + 4H₂O; coefficients 1, 5, 3, 4

Ba²⁺(aq) + SO₄²⁻(aq) → BaSO₄(s); correct species and charges; correct state symbols

spectator ions appear unchanged on both sides of the equation; omitting them shows only the species that actually undergo chemical change

the weighted mean mass of an atom of an element; compared with 1/12 the mass of a ¹²C atom

Relative formula mass of calcium nitrate, Ca(NO₃)₂, using Ar values Ca = 40.1, N = 14.0, O = 16.0.

The formula Ca(NO₃)₂ contains one Ca atom and two nitrate (NO₃) groups, giving 1 Ca, 2 N, and 6 O in total.

Mr = Ar(Ca) + 2 × [Ar(N) + 3 × Ar(O)] = 40.1 + 2 × (14.0 + 3 × 16.0) = 40.1 + 2 × (14.0 + 48.0) = 40.1 + 2 × 62.0 = 40.1 + 124.0 = 164.1

Check by summing individual atoms: Ca (1 × 40.1 = 40.1) + N (2 × 14.0 = 28.0) + O (6 × 16.0 = 96.0) = 164.1.

Relative formula mass = 164.1 (no units, since Mr is a ratio).

NaCl exists as a giant ionic lattice, not as discrete molecules; the term 'molecular mass' would therefore be inaccurate

Step 1 — Definition of ppm by mass: ppm = (mass of component ÷ total mass of mixture) × 10⁶. Step 2 — Substitute (mass NO₂ = 1.6 × 10⁻³ g, total air = 50 g): ppm = (1.6 × 10⁻³ ÷ 50) × 10⁶. Step 3 — Mass fraction: 1.6 × 10⁻³ ÷ 50 = 3.2 × 10⁻⁵. Step 4 — × 10⁶: 3.2 × 10⁻⁵ × 10⁶ = 32. Final answer: concentration of NO₂ = 32 ppm.

Volume conversion (cm³ to dm³) $$V=\frac{500}{1000}=0.500{\text{ dm}}^{3}$$ Finding the concentration $$c=\frac{n}{V}=\frac{0.250}{0.500}=0.500{\text{ mol dm}}^{-3}$$

Finding moles of NaCl $$n=\frac{m}{{M}_{r}}=\frac{5.85}{58.5}=0.100\text{ mol}$$ Volume conversion (cm³ to dm³) $$V=\frac{200}{1000}=0.200{\text{ dm}}^{3}$$ Finding the concentration $$c=\frac{n}{V}=\frac{0.100}{0.200}=0.500{\text{ mol dm}}^{-3}$$

0.250 x 180.0 = 45.0 g/dm3

Finding the molar mass of Na₂CO₃ $${M}_{r}({\text{Na}}_{2}{\text{CO}}_{3})=(2\times 23.0)+12.0+(3\times 16.0)=106.0{\text{ g mol}}^{-1}$$ Finding moles of Na₂CO₃ $$n=\frac{2.00}{106.0}=0.01887\text{ mol}$$ Volume conversion (cm³ to dm³) $$V=\frac{250}{1000}=0.250{\text{ dm}}^{3}$$ Finding the concentration in mol dm⁻³ $$c=\frac{n}{V}=\frac{0.01887}{0.250}=0.0755{\text{ mol dm}}^{-3}$$ Finding the concentration in g dm⁻³ $${c}_{{\text{g dm}}^{-3}}=0.0755\times 106.0=8.00{\text{ g dm}}^{-3}$$ Assumption: the solute fully dissolves and the total solution volume is exactly 0.250 dm³ at the calibration temperature.

Finding moles of each element $${n}_{C}=\frac{40.0}{12.0}=3.33$$ $${n}_{H}=\frac{6.7}{1.0}=6.70$$ $${n}_{O}=\frac{53.3}{16.0}=3.33$$ Ratio C:H:O = 1:2:1, so the empirical formula is CH₂O. Finding the multiplier $${M}_{r}(\text{empirical})=30$$ $$n=\frac{180}{30}=6$$ The molecular formula is C₆H₁₂O₆.

Empirical formula gives the simplest whole-number ratio of atoms in a compound; molecular formula gives the actual number of atoms of each element in one molecule.

$$n(\text{Mg})=\frac{4.86}{24.3}=0.200\text{ mol}$$ $$n(\text{MgO})=0.200\text{ mol (1:1 ratio)}$$ $$m(\text{MgO})=0.200\times 40.3=8.06\text{ g}$$