the Empirical formula of the compound is calculated as follows
find the moles of each element
moles = mass/molar mass
that is Magnesium= 0.564 g /24 g/mol = 0.0235 moles
For oxygen = 0.376 g /16 g/mol= 0.0235 moles
find the mole ratio by dividing both moles of by 0.0235 moles
magnesium = 0.0235/0.0235= 1mole
oxygen= 0.0235 /0.0235= 1mole
therefore the empirical formula = MgO
Answer:
C₄H₈O₂.
Explanation:
- Firstly, we can calculate the no. of moles (n) of each component using the relation:
<em>n = mass/atomic mass,</em>
mol C = mass/(atomic mass) = (54.5 g)/(12.0 g/mol) = 4.54 mol.
mol H = mass/(atomic mass) = (9.3 g)/(1.0 g/mol) = 9.3 mol.
mol O = mass/(atomic mass) = (36.2 g)/(16.0 g/mol) = 2.26 mol.
- To get the empirical formula, we divide by the lowest no. of moles (2.26 mol) of O:
∴ C: H: O = (4.54 mol/2.26 mol) : (9.3 mol/2.26 mol) : (2.26 mol/2.26 mol) = 2: 4: 1.
<em>∴ Empirical formula mass of (C₂H₄O) = 2(atomic mass of C) + 4(atomic mass of H) + 1(atomic mass of O) =</em> 2(12.0 g/mol) + 4(1.0 g/mol) + (16.0 g/mol)<em> = 44.0 g/mol.</em>
∴ Number of times empirical mass goes into molecular mass = (88.0 g/mol)/(44.0 g/mol) = 2.0 times.
∴ The molecular formula is, 2(C₂H₄O), that is; <em>(C₄H₈O₂)</em>
The computation for this problem is:
(1.55x10^4 / 1.0x10^3) x 19.8 mm Hg
= 15.5 x 19.88 mm Hg
= 308.14 mm Hg decrease
= 308.14 x 0.05 C = 15.407 deg C
deduct this amount to 100
100 – 15.407 = 84.593 C
ANSWER: 85 deg C (rounded to 2 significant figures)
