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.

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

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LucianoBordoli LucianoBordoli · Answer 2 · 2019-12-09T20:27:29+01:00

Answer:

[tex]0.017602moles[/tex]

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 :

[tex]Cu=63.546\frac{g}{mole}[/tex]

[tex]Br=79.904\frac{g}{mole}[/tex]

[tex]H=1.00784\frac{g}{mole}[/tex]

[tex]O=15.999\frac{g}{mole}[/tex]

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

(CuBr2•4(H2O)) ⇒

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

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 :

[tex]\frac{295.41272g}{1mole}=\frac{5.2g}{x}[/tex]

Solving for x :

[tex]x=\frac{(5.2g).(1mole)}{295.41272g}=0.017602moles[/tex]

A total of [tex]0.017602moles[/tex] were measured.