The molecular formula of organic solvent is <em>C6H12</em>
<h2>calculation</h2><h3>find the empirical formula first as in step 1 and 2</h3>
Step 1: f<em>ind the moles of C and H</em>
- moles = % composition/molar mass
- from periodic table molar mass of C= 12 g/mol while that of H= 1 g/mol
- moles is C is therefore = 85.6/12= 7. 13 moles
- moles of H= 14.4/1 - 14.4 moles
Step 2: <em>calculate the mole fraction by dividing each mole by smallest number of mole(7.13)</em>
H= 14.4/7.13 =2
the empirical formula is therefore = CH2
<h2>Then calculate the molecular formula from empirical formula</h2>
step 3: divide the grams molar mass by empirical formula mass
empirical formula mass = 12+(1 x2) = 14 g/mol
= 84.2/ 14 = 6
step 4: multiply each of the subscript within the empirical formula with the value gotten in step 3
- that is [CH2]6 = C6H12 therefore the molecular formula = <u>C6H12</u>
Yo sup??
Let the percentage of K-39 be x
then the percentage of K-40 is 100-(x+0.01)
We know that the net weight should be 39.5. Therefore we can say
(39*x+40*(100-(x+0.01))+38*0.01)/100=39.5
(since we are taking it in percent)
39*x+40*(100-(x+0.01))+38*0.01=3950
39x+4000-40x-0.4+0.38=3950
2x=49.98
x=24.99
=25 (approx)
Therefore K-39 is 25% in nature and K-40 is 75% in nature.
Hope this helps.
Answer:
27 g
Explanation:
M(C6H12O6) = 6*12 + 12*1 + 6*16 = 180 g/mol
100 mL = 0.1 L solution
1.5 M = 1.5 mol/L
1.5 mol/L * 0.1 L = 0.15 mol C6H12O6
0.15 mol * 180 g/1 mol = 27 g C6H12O6
