How many moles of calcium chloride (CaCl2) are required to react completely with 1.00 mole of citric acid?

answer is: empirical formula is tl₂o. if we use 100 grams of compound, tha:

m(tl) = 96.2%  · 100 g ÷ 100%.

m(tl) = 96.2 g.

n(tl) = m(tl)  ÷ m(tl). n(tl) = 96.2 g  ÷ 204.4 g/mol.

n(tl) = 0.47 mol. m(o) = 3.77 g. n(o) = m(o)  ÷ m(o). n(o) = 3.77 g  ÷ 16 g/mol. n(ag) = 0.235 mol. n(tl) . n(o) = 0.47 mol : 0.235 mol /0.235. n(tl) . n(o) = 2 : 1.

m(tl₂o) = 204.4 g/mol  · 2 + 16 g/mol = 424.8 g/mol.

m(compound) = 456. 8 g/mol - 424.8 g/mol = 32 g/mol ( 2 oxygens); tl₂o₃. 

5 structural isomers

the isomers are:

n-exane :   ch3ch2ch2ch2ch2ch3

2-methylpentane :   ch3ch(ch3)ch2ch2ch3

3-methylpentane :   ch3ch2ch(ch3)ch2ch3

2,3-dimethylbutane :   (ch3)2chch(ch3)2

2,2-dimethylbutane :   (ch3)3cch2ch3