Question

Urgent!! A chemist measured 5.2 g copper(II) bromide tetrahydrate (CuBr2•4(H2O)). How many moles were measured out? Answer in units of mole.

Answer 1

Answer:

$0.017602moles$

Explanation:

The first step is to find the weight per mol of (CuBr2•4(H2O))

We can do this replacing the elements Cu,Br,H,O by their respectively molar masses.

The molar mass of an element represents its weight per mole of element.

For the elements :

$Cu=63.546\frac{g}{mole}$

$Br=79.904\frac{g}{mole}$

$H=1.00784\frac{g}{mole}$

$O=15.999\frac{g}{mole}$

Replacing this values in the copper(II) bromide tetrahydrate equation

(CuBr2•4(H2O)) ⇒

$63.546+(79.904).2+4.[2(1.00784)+15.999]=295.41272$

We find that a mole of copper(II) bromide tetrahydrate has a mass of 295.41272 g.

Given that we have 5.2 g in the sample, we can calculate the total moles as :

$\frac{295.41272g}{1mole}=\frac{5.2g}{x}$

Solving for x :

$x=\frac{(5.2g).(1mole)}{295.41272g}=0.017602moles$

A total of $0.017602moles$ were measured.

Answer 2

  The  moles  which  were   measured  out  is  calculated  using  the  following  formula

moles  =  mass/molar   mass

molar mass  of  CuBr2.4H20  =   63.5  Cu + (  2  x79.9)  br  + ( 18  x4_)  h20  =  295.3  g/mol

moles  is therefore=  5.2 g/  295.3 g/mol=  0.0176 moles