What are you doing Sat 14 Apr? Want to come down for a surf safari party?

C(H₂O) = 4.18 J/g/°C
m(H₂O) = 2.46 kg = 2460 g
ΔT = 27.31°C - 25.24°C = 2.07°C
Q = C × ΔT × m = 4.18 J/g/°C × 2.07°C × 2460 g = 21285 J = 21.3 kJ

C(Cu) = 0.385 J/g/°C
m(Cu) = 30.0 g
ΔT = Q ÷ C ÷ m = 250J ÷ 0.385J/g/°C ÷ 30.0g = 21.6°C
Final temperature = 22.0°C + 21.6°C = 43.6°C ✓

ΔT = 25°C - 15°C = 10°C
C(H₂O) = 4.18 J/g/°C
m(H₂O) = 175 g
Q = C × ΔT × m = 4.18J/g/°C × 10°C × 175g = 7320 J = 7.32 kJ ✓

ΔU = +300 J + +700J = +1000 J ✓
System is endothermic as it has absorbed heat. ✓

M(Ca) = 40.08 g/mol
M(P) = 30.97 g/mol
M(O) = 16.00 g/mol
∴ M(Ca₃(PO₄)₂) = 3 × 40.08 + 2 × 30.97 + 8 × 16.00 = 310.18 g/mol

M(Na) = 22.99 g/mol
M(H) = 1.01 g/mol
M(C) = 12.01 g/mol
M(O) = 16.00 g/mol
∴ M(NaHCO₃) = 22.99 + 1.01 + 12.01 + 3 × 16.00 = 84.01 g/mol

In a 100g sample, m(C) = 14.5g and m(Cl) = 85.5g
n(C) = m(C) ÷ M(C) = 14.5g ÷ 12.01g/mol = 1.21 mol
n(Cl) = m(Cl) ÷ M(Cl) = 85.5g ÷ 35.45g/mol = 2.41 mol
Molar ratio = 1.21 mol C : 2.41 mol Cl
= 1 : 1.99
≈ 1 : 2
Empirical formula = CCl₂
Molar ratio is 1:2, Empirical formula is CCl₂

m(O) = 0.896g - 0.111g - 0.477g = 0.308g
n(Na) = m(Na) ÷ M(Na) = 0.111g ÷ 22.99g/mol = 0.00483 mol
n(Tc) = m(Tc) ÷ M(Tc) = 0.477g ÷ 98.9g/mol = 0.00482 mol
n(O) = m(O) ÷ M(O) = 0.308g ÷ 16.0g/mol = 0.0192 mol
Molar ratio = 0.00483 mol Na : 0.00483 mol Tc : 0.0192 mol O
= 1 : 1 : 3.986
≈ 1 : 1 : 4
Empirical formula = NaTcO₄
If the question asks for a formula, you must answer with a formula!

(a) Ca(OH)₂ + 2HCl → CaCl₂ + 2H₂O ✓
(b) 2AgNO₃ + CaCl₂ → Ca(NO₃)₂ + 2AgCl ✓
(c) 2Fe₂O₃ + 3C → 4Fe + 3CO₂ ✓
(d) 2NaHCO₃ + H₂SO₄ → Na₂SO₄ + 2H₂O + 2CO₂ ✓
(e) C₄H₁₀ + 6½O₂ → 4CO₂ + 5H₂O
∴ 2C₄H₁₀ + 13O₂ → 8CO₂ + 10H₂O

(a) disulphur dichloride ✓
(b) iodine heptafluoride ✓
(c) hydrogen bromide ✓
(d) dinitrogen trioxide ✓
(e) silicon carbide ("carboxide" also has oxygen)
(f) methanol ✓

(a) CH₄ ✓
(b) HI ✓
(c) CaH₂ (not ClH = HCl = Hydrogen chloride = Hydrochloric acid)
(d) PCl₃ ✓
(e) N₂O₅ (not N₂H₅)
(f) SF₆ (not SF₈)
(g) BF₃ ✓

(a) pentane (or sometimes n-Pentane) ✓
(b) 2-methylpentane ✓
(c) 2,4-dimethylhexane ✓

(a) Has a C≡C (carbon-carbon triple bond), therefore is an Alkyne.
(b) Has a C=O (carbonyl group) on end carbon (ie -CHO or formyl group), therefore is an Aldehyde. ✓
(c) Has a C=O (carbonyl group), on a non-end carbon, therefore is a Ketone. ✓

(a) Has a C=C (carbon-carbon double bond), therefore is an Alkene. ✓
(b) Has a -COOH (carboxyl group), therefore is a Carboxylic Acid.
(c) Has a -OH (hydroxyl group), therefore is an Alcohol. ✓
(d) Has a -OH (hydroxyl group), therefore is an Alcohol. ✓

From Q 1.37:
4.665g N : 1.00g H
= 4.56g N : 0.977g H
m(ammonia) = 4.56g + 0.977g = 4.64g

✓ Beauty!

4.12g Cl : 1.20g P
= 1.00g Cl : 0.291g P
= 6.22g Cl : 1.81g P
Answer: 1.81g P

✓ Beauty!

2.00g H : 9.33g N
= 1.00g H : 4.665g N
= 6.28g H : 29.30g N
Answer: 29.3g N

✓ Beauty!

(a) 3.65g Ca : 4.13g Cl = 1.00g Ca : 1.13g Cl
(b) 0.856g Ca : 1.56g Cl = 1.00g Ca : 1.82g Cl
(c) 2.45g Ca : 4.57g Cl = 1.00g Ca : 1.87g Cl
(d) 1.35g Ca : 2.39g Cl = 1.00g Ca : 1.77g Cl
(e) 5.64g Ca : 9.12g Cl = 1.00g Ca : 1.62g Cl
Therefore (d) is the correct answer.

✓ Unlike the first one, these are much closer together, so you can't take a shortcut.

(a) 6.35g N : 7.26g O = 0.875g N : 1.00g O
(b) 4.63g N : 10.58g O = 0.438g N : 1.00g O
(c) 8.84g N : 5.05g O = 1.75g N : 1.00g O
(d) 9.62g N : 16.5g O = 0.583g N : 1.00g O
(e) 14.3g N : 40.0g O = 0.358g N : 1.00g O
Therefore (c) is the correct answer.

✓ As you mention, m(N) needs to be bigger than m(O), so you can quickly jump to the correct answer.

Here we go again! :D

Yeah i know what you mean :( never enough hours in the day to get in the gaming/other things one needs to do.

Man that Orbitals guide is awesome :D

Actually, in the last two weeks, I was working. I had an old friend ring up out of the blue for… get this… some contract mathematics!
I'd just three days earlier made a comment that there wasn't much cash in being a Mathematician For Hire.

Besides that, I'm surfing at least 4 times a week. Things are pretty cruisy! Although I'm still pining for Azeroth - I had to go play Warcraft III just to get a fix.

Watch this space! I'm putting something together for you: Atomic Orbitals 1

It should be as easy as 1.0 [E] 14 [-] [ / ] 4.4 [E] 8 [-] [=]

Oh no man i hope you is ok! I have just been doing notes so its all good! Only 14 chapters left huzza!

After exams, I got sick :(

If you haven't worked it out yet…
Scientific calculators should have an "E" button. So, to put in 4.4 x 10^-8, you type:
4.4 [E] 8 [-]

Logs need a log button. If you have a scientific calculator, there should be a [log] and a [ln]. Use the [log] for pH calcs.

example of a expo q is 1.0 x 10^-14/4.4 x 10^-8 kind of thing. The majority of the pH questions i am guessing are -log's as well

Hope your well and work has settled down a little!

… doubly so because I've been sick, had my car break down, am up to my eyeballs in these bloody assignments, and have a wedding tomorrow.

I'll be calmer come Tuesday (4/10)… and a lot calmer come Sat 22/10 (exams will be over by then).

Example formula: Mg _(s) + 2HCl _(aq) → Mg²⁺ _(aq) + 2Cl^- _(aq) + H_{2 (g)}

You can go up and down as much as you like (assuming the quantity exists/makes sense), but you can only go sideways using moles. Using this idea, you can "track a path" of calculations necessary to get from what you've got to what you're trying to find.

In a 100 g sample, m(Na) = 32.4 g, m(S) = 22.6g, and m(O) = 45.0 g
n(Na) = m(Na) ÷ M(Na) = 32.4 ÷ 22.99 = 1.41 mol
n(S) = m(S) ÷ M(S) = 22.6 ÷ 32.07 = 0.70 mol
n(O) = m(O) ÷ M(O) = 45.0 ÷ 16.00 = 2.81 mol
Mole ratio is 1.41 : 0.70 : 2.81 = $\frac{1.41}{0.70} : \frac{0.70}{0.70} : \frac{2.81}{0.70}$ = 2 : 1 : 4
Therefore, the empirical formula is Na₂SO₄

✓ Ripper!

In a 100 g sample, m(Ba) = 69.6 g, m(C) = 6.09g, and m(O) = 24.3 g
n(Ba) = m(Ba) ÷ M(Ba) = 69.6 ÷ 137.3 = 0.507 mol
n(C) = m(C) ÷ M(C) = 6.09 ÷ 12.01 = 0.507 mol
n(O) = m(O) ÷ M(O) = 24.3 ÷ 16.00 = 1.52 mol
Mole ratio is 0.507 : 0.507 : 1.52 = $\frac{0.507}{0.507} : \frac{0.507}{0.507} : \frac{1.52}{0.507}$ = 1 : 1 : 3
Therefore, the empirical formula is BaCO₃

✓ Total blitz!

For 1 mole of N₂O₄:
m(N₂O₄) = M(N₂O₄) = 2 × 14.01 + 4 × 16.00 = 92.02 g
m(N) = 2 × M(N) = 2 × 14.01 = 28.02 g
m(O) = 4 × M(O) = 4 × 16.00 = 64.00 g
%m(N) = m(N) ÷ m(N₂O₄) = 28.02 ÷ 92.02 = 30.45%
%m(O) = m(O) ÷ m(N₂O₄) = 64.00 ÷ 92.02 = 69.55%

✓ All good.

M(N₂O₅) = 2 × 14.01 + 5 × 16.00 = 108.02 g mol^-1
M(2N) = 2 × 14.01 = 28.02 g mol^-1
M(5O) = 5 × 16.00 = 80.00 g mol^-1
%m(N) = m(N) ÷ m(N₂O₅) = M(2N) ÷ M(N₂O₅) = 28.02 ÷ 108.02 = 25.94%
%m(O) = m(O) ÷ m(N₂O₅) = M(5O) ÷ M(N₂O₅) = 80.00 ÷ 108.02 = 74.06%

✓ All good… so far. The pedantic marker would say you haven't answered the question asked, so to be safe, you should add:

Thus yes, mass % of 25.94% N and 74.06% O are consistent with a molecular formula of N₂O₅.

Because mole ratio = 2:3, must be 6g as 6 + 9 = 15 and is in a ratio of 2:3.

✗ Like the previous question, the 2:3 ratio is a mole ratio, not a mass ratio.

M(Fe₂O₃) = 2 × 55.85 + 3 × 16.00 = 159.7 g mol^-1
n(Fe₂O₃) = m ÷ M = 15 ÷ 159.7 = 0.09393 mol
n(Fe) = 2 × n(Fe₂O₃) = 2 × 0.09393 mol = 0.1879 mol
m(Fe) = n(Fe) × M(Fe) = 0.1879 × 55.85 = 10.5 g

Ratio of Fe:O = 2:3
m(Fe) = 2/3 × m(O) = 2/3 × 25.6 g = 17.07 g

✗ It looks like you were on the right track first up, but crossed that out. The ratio 2:3 is a molar ratio, not a mass ratio. However, your answer would have been correct if an atom of iron weighed the same as an atom of oxygen.

n(O) = m(O) ÷ M(O) = 25.6 g ÷ 16.00 g mol^-1 = 1.6 mol
n(Fe) = n(O) × 2 ÷ 3 = 1.067 mol
m(Fe) = n(Fe) × M(Fe) = 55.85 × 1.067 = 59.57 g

n = m ÷ M = 0.0011 ÷ 24.31 = 4.5×10^-5
Mole ratio of C:Mg = 55:1
n(C) = 2.5×10^-3
m(C) = M × n = 12.01 × 2.5×10^-3 = 0.030 g

✓ All good!

M(H₂SO₄) = 2 × 1.01 + 1 × 32.07 + 4 × 16.00 = 98.09 g mol^-1
n = m ÷ M = 45.8 ÷ 98.09 = 0.47 mol

✓ All good!

M(Na₂CO₃) = 2 × 22.98 + 1 × 12.01 + 3 × 16.00 = 105.97 g mol^-1
m = M × n = 105.97 × 0.125 = 13.25 g

✓ All good!

M(Ca₃(PO₄)₂) = 3 × 40.08 + 2 × 30.97 + 8 × 16.00 = 310.2 g mol^-1
m = M × n = 310.2 × 0.115 = 35.7 g

✓ All good!

M(S) = 32.06 g mol^-1
n(S) = m ÷ M = 35.6 g ÷ 32.06 g mol^-1 = 9.087 mol

✗ Missed by that much! The problem here is that we're interested in S₈, not S. So…

M(S) = 32.06 g mol^-1
M(S₈) = 8 × 32.06 = 256.48 g mol^-1
n = m ÷ M = 35.6 g ÷ 256.48 g mol^-1 = 0.139 mol

n = m ÷ M = 109.13 g ÷ 12.01 g mol^-1 = 9.087 mol

✓ All good!

%q(Part) = q(Part) ÷ q(Whole) × 100%

where q is any measurable quantity.

%m(N) = m ÷ m = 0.1417 ÷ 0.5462 × 100 = 25.94%
%m(O) = m ÷ m = 0.4045 ÷ 0.5462 × 100 = 74.06%

✓ Yep. Easy as! God stuff!

% by mass is:
%m(C) = m ÷ m = 5.217 ÷ 8.657 × 100 = 60.25%
%m(H) = m ÷ m = 0.962 ÷ 8.657 × 100 = 11.11%
%m(O) = m ÷ m = 2.478 ÷ 8.657 × 100 = 28.62%

✓ Yep all good. Although, the question should specify % composition by mass, 'cos you could do % composition by molarity:

% by moles is:
n(C) = m ÷ M = 5.217 ÷ 12.01 = 0.4344 mol
n(H) = m ÷ M = 0.962 ÷ 1.008 = 0.9544 mol
n(O) = m ÷ M = 2.478 ÷ 16.00 = 0.1549 mol
%n(C) = 0.4344 ÷ (0.4344 + 0.9544 + 0.1549) × 100 = 28.14%
%n(H) = 0.9544 ÷ (0.4344 + 0.9544 + 0.1549) × 100 = 61.83%
%n(O) = 0.1549 ÷ (0.4344 + 0.9544 + 0.1549) × 100 = 10.03%

By mass is common, but not very useful from a chemical perspective. By moles is less common, but more useful. Bit twisted, hey?