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Relative masses of atoms and molecules Notes igcse

Relative masses of atoms and molecules Notes for iGCSE Chemistry Notes CIE prepared by iGCSE Teachers

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Relative masses of atoms and molecules Notes for iGCSE

Core Syllabus

  • Describe relative atomic mass, Ar , as the average mass of the isotopes of an element compared to 1/12th of the mass of an atom of 12C
  •  Define relative molecular mass, Mr , as the sum of the relative atomic masses. Relative formula mass, Mr , will be used for ionic compounds
  •  Calculate reacting masses in simple proportions. Calculations will not involve the mole concept

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GEOMETRIC SEQUENCE….

  • THE DEFINATION..

I give you the first term of a sequence, say \(u_1\) =5 and  I ask you to multiply by a fixed number, say r =2, in order to find the next term. The following sequence is generated:

5, 10, 20, 40, 80, …

Such a sequence is called geometric. That is, in a geometric sequence the ratio between any two consecutive terms is constant.

In other words , we can say , geometric sequence is a sequence where there is a common ratio between consecutive number.

For determining geometric sequence , we only need 

  • The first term
  • The common ratio
  • Common ratio:  The common ratio is the constant factor by which each term in a geometric sequence is multiplied to get the next term.

EXAMPLE 1
(a) \(u_1\) =1,  r =2        the sequence is 1, 2, 4, 8, 16, 32, 64, …

(b) \(u_1\) =5,  r = 10    the sequence is 5, 50, 500, 5000, …

(c) \(u_1\) =1,  r = -2      the sequence is 1,-2, 4,-8, 16, …

(d) \(u_1\) =1, r = \(\frac{1}{2}\)    the sequence is  1, \(\frac{1}{2}\), \(\frac{1}{4}\), \(\frac{1}{16}\), \(\frac{1}{32}\), …..

(e) \(u_1\) =1, r = \(\frac{-1}{2}\)    the sequence is 1, \(\frac{-1}{2}\), \(\frac{1}{4}\), \(\frac{1}{16}\), \(\frac{-1}{32}\), …..

ATTENTION!!

  • The common ratio r may also be negative! In this case the signs alternate (+, -, +, -, …) [see (c) and (e) above].
  • The common ratio r may be between -1 and 1, that is |r|<1. In such a sequence the terms approach 0 [see (d) and (e) above]
  • THE \(N^{th}\) TERM FORMULA

The nth term formula for a geometric sequence is \(u_n =u_1r^{n-1}\)

where,

  • \(u_1\) is the first term
  • n is the term number
  • r is the common ratio

Indeed, let us think:
In order to find \(u_5\) , we start from u1 and then multiply 4 times by the ratio r

Hence, \(u_5 = u_1r^4\)

Similarly, \(u_{10} = u_1r^9\), \(u_{50} = u_1r^{49}\) and so on…..

EXAMPLE 2
In a geometric sequence let \(u_1\) =3 and r =2. Find
(a) the first four terms

(b) the 100th term

Solution
(a) 3, 6, 12, 24
(b) Now we need the general formula

\(u_{100} = u_1r^{99} = 3.2^{99}\)

EXAMPLE 3
In a geometric sequence let \(u_1\) =10 and \(u_{10} = 196830 . Find \(u_3\)

Solution
We know \(u_1\) , we need r. We exploit the information for \(u_{10}\) first.

\(u_{10}=u_1r^9 = 196830 = 10 . r^9\)

\(r^9= 19683\)

\(r= \sqrt[9]{19683} = 3\)

Therefore , \(u_3 = u_1r^2 = 10.3^2 = 90\)

REMEMBER!!

Our first task in a G.S. is to find the basic elements, \(u_1\) and r , and then anything else.

  • SUM OF N TERMS(\(S_n\)

Given that r ≠ 1 , the result is given by

\(S_n = \frac{u_1(r^n-1)}{r-1}\) or  \(S_n = \frac{u_1(1-r^n)}{1-r}\)

 

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