Some formula:
Relative isotopic mass = mass of 1 atom of the isotope / 1/12 the mass of a C12 atom
Relative atomic mass, Ar =
weighted average mass of an atom of the element / 1/12 the mass of a C12 atom
Question:
Chlorine has 2 isotopes. 35Cl and 37Cl in the ratio 3:1 in abundancies. What is the relative atomic mass of Chlorine?
An example question from my lecture notes... I don't kinda understand the correlation between relative isotopic mass and relative atomic mass. Need help urgently, thanks!
Cl-35 and Cl-37 are isotopes of chlorine.
You need to know the definition of Isotope. Basic definition is.. atoms of the same element with same proton but different number of neutrons.
The Ar, relative atomic mass, is calculate by using the average mass of the isotopes with relation to the abundances.
Calculation wise.
(35 x 0.75) + (37 x 0.25)
Try this question to reinforce your knowledge by applying what you have learnt.
Q: Naturally occurring silver is 51.84% Ag-107 and 48.16% Ag-109, calculate the relative atomic mass of silver.
(question from http://www.ausetute.com.au/atomicmass.html)
Notice your new periodic table in A level is slightly different that was used in O level. :) You will see more decimals for the atomic mass for element. Yep, that is because it involves more precise calculation.
Using this knowledge, you can then work backwards to find the relative abundances of the elements. By using simple algebra...
Hi! I do have questions on stoichiometry too. But before that, may I ask, what's the meaning of "1/12 the mass of a C12 atom"? Is it '1' since 1/12 of 12 is 1? And what is it TS's formula cannot be used to solve the question? Darkness_hacker99's method is similar to that used in Secondary syllabus and is the 3:1 ratio compared to a value of 100%? That's why it's 0.75 and 0.25.
Ok my questions...
Which of the following contains 1 mole of the stated particles?
A Electrons in 1g of hydrogen gas
B Ions in 100g of calcium carbonate
D Oxygen atoms in 1 mole of oxygen gas ( i know it's obviously wrong but is there any way to prove that it's wrong? i mean, how do i present workings to validate my claim?)
Answer being A, I chose B. Why is it that the reference to 'electrons' is accepted? For option B, am I wrong because CaCO3 is insoluble and thus cannot dissociate into ions? (I think it's this case, hahaha)
2. Which statement(s) about a 12.0g sample of 12C are correct?
There is this statement which I marked as 'correct' when in fact it is wrong...
"The no. of atoms is the same as the number of atoms in 2.0g of 1 H2..." (1 is written as a top subscript)
I mean the no. of moles of H atoms will be = 2.0 / (1+1)
=1 mol
So isn't no. of moles = 1 X 6.02 X 10^23, same as that of 12.0g sample of 12C?
3. Lastly
How many moles of fluoride ions are there in 20g of aluminium fluoride?
my workings: no of moles of AlF3 = 20 / (19 X 3 + 27) = 0.238095 mol
so no. of fluoride ions = 0.238095 X (3/4)
= 0.179 mol (3 s.f)
As to why I multiply by 3/4, well aluminium fluoride is made up of the ions
Al3+ + 3F- .... a total of 4 units. that's how i kind of worked it out, though i'm not sure of the accuracy of this method.
Thanks
>>> Which of the following contains 1 mole of the stated particles?
A Electrons in 1g of hydrogen gas
B Ions in 100g of calcium carbonate
D Oxygen atoms in 1 mole of oxygen gas ( i know it's obviously wrong but is there any way to prove that it's wrong? i mean, how do i present workings to validate my claim?)
Answer being A, I chose B. Why is it that the reference to 'electrons' is accepted? For option B, am I wrong because CaCO3 is insoluble and thus cannot dissociate into ions? (I think it's this case, hahaha) <<<
B is wrong because you only considered moles of the ionic compound, not moles of ions within the compound. You have to consider both cations and anions. Just because CaCO3 is insoluble in water does not mean cations and anions are not present. Just because you can't separate your hands from your body does not mean they are not present and that you needn't count them.
There are two moles of O atoms per mole of O2 molecules.
>>> Which statement(s) about a 12.0g sample of 12C are correct?
There is this statement which I marked as 'correct' when in fact it is wrong...
"The no. of atoms is the same as the number of atoms in 2.0g of 1 H2..." (1 is written as a top subscript)
I mean the no. of moles of H atoms will be = 2.0 / (1+1)
=1 mol
So isn't no. of moles = 1 X 6.02 X 10^23, same as that of 12.0g sample of 12C? <<<
No. of atoms in 12g sample of C = 1 mol of atoms
No. of atoms in 2g of H2 = 1 mol of H2 molecules = 2 mol of H atoms (since 1 mole of H2 molecules contain 2 moles of H atoms)
>>> How many moles of fluoride ions are there in 20g of aluminium fluoride?
my workings: no of moles of AlF3 = 20 / (19 X 3 + 27) = 0.238095 mol
so no. of fluoride ions = 0.238095 X (3/4) = 0.179 mol (3 s.f) <<<
No. of F- ions = no. of mol of AlF3 x 3 = 0.238095 x 3 = 0.714 mol
Thanks UltimaOnline for the clear explanations :)
I've 2 questions which I'm unsure of. So, if there's anyone who can solve them, feel free to help me out.
1. When 1.01g sample of potassium nitrate is heated above its m.p, oxygen gas is evolved. The mass of the sample decreases by 0.16g, suggest the identity for the residue and hence construct a balanced equation for this decomposition reaction.
[K = 39.1; N = 14.0; O = 16.0]
I'd guess the identity of residue to be potassium nitride but I don't know how I get prove this with workings. (this hasn't been taught yet but I just wanna try out the question and see the steps involved in solving it)
2. Sulfur and chlorine can react together to fom S2Cl2. When 1.00g of this sulfur chloride reacted with water, 0.36g of a yellow ppt of sulfur was formed together with a solution containing a mixture of sulfurous acid, H2SO3, and HCl acid. Use the above data to deduce the equation for the reaction between S2Cl2 and water. [ S =32.1; Cl = 35.5]
'O' & 'A' Level Qns.
Qn 1.
When 1.01g sample of potassium nitrate(V) is heated above its melting point, oxygen gas is evolved. The mass of the sample decreases by 0.16g. Construct a balanced equation for this decomposition reaction, name the residue and (for 'A' level students) draw its structural formula.
Q2.
Sulfur and chlorine can react together to form disulfur dichloride. When 1.00g of disulfur dichloride was reacted with water, 0.36g of a yellow precipitate of sulfur was formed together with a solution containing a mixture of sulfuric(IV) acid (a.k.a. sulfurous acid), and hydrochloric acid. Given that the most common isotope of sulfur is S8, use the above data to deduce the balanced equation for the reaction between disulfur dichloride and water. (For 'A' level students) draw the structural formulae of disulfur dichloride and sulfuric(IV) acid.
Solutions (partial).
Qn1. Since 1 mole of potassium nitrate(V) decomposes to give 0.5 mole of oxygen gas, the residue is potassium nitrate(III) {stock name} or potassium nitrite {latin name}.
For the nitrate(III) ion (aka nitrite ion), uninegative formal charge on a singly bonded O atom; central N atom has 1 lone pair, 3 bond pairs; electron geometry tetrahedral, ionic geometry trigonal pyramidal.
Qn2. Since 1 mole of disulfur dichloride undergoes hydrolysis to give 1.5 moles of S(s), or 1.5/8 moles of S8(s), the coefficients of the balanced equation are : 2, 3, 1, 4, 3/8
For disulfur dichloride, the 2 central S atoms have 2 lone pairs and 2 bond pairs; electron geometry tetrahedral, molecular geometry v-shape.
For sulfuric(IV) acid {stock name} or sulfurous acid {latin name}, the central S atom has 1 lone pair, 4 bond pairs (S can expand its octet using its vacant and energetically accessible 3d orbitals); electron geometry tetrahedral, molecular geometry trigonal pyramidal.