المشاركة الأصلية كتبت بواسطة cool-cat يانجوول

Practice 1.
• A compound contains C, H, N. Combustion of 35.0mg of the compound produces 33.5mg CO2 and 41.1mg H2O. What is the empirical formula of the compound?
• Solution:
• 1. Determine C and H, the rest from 33.5 mg is N.
• 2. Determine moles from masses.
• 3. Divide by smallest number of moles
all the C ends up as CO2
mg of C = (12.011/44.011) x 33.5 mg CO2 = 9.142 mg C
all of the H ends up as H2)
mg H = (2.016/18.016) x 41.1 mg H2O = 4.599 mg H
mg N = 35.0 - 9.14 - 4.60 = 21.26
CxHyNz represents the empirical formula
where x = moles of C atoms, y = moles of H atoms, and z = moles of N atoms in (1 mole of) the molecule
Calculate the moles of C, H, and N that you found.
Determine the ratio of the moles x : y : z
Do that and then look at my calculations-
9.14 mg C / 12.011 mg/mmol = 0.76 millimol C
4.60 mg H / 1.008 mg/mmol = 4.60 millinol H
21.26 mg N / 14.006 mg/mmol = 1.52 millimol N
divide by the smallest number (0.76)
C = 1
H = 6
N = 2
CH6N2
Practice 2.
• Caffeine contains 49.48% C, 5.15% H, 28.87% N and 16.49% O by mass and has a molar mass of 194.2 g/mol. Determine the molecular formula.
Solution:
1.Always assume Total mass of the sample is 100 g.
2.Convert mass to moles.
3.Divide all mole values by the lowest value of mole
4.Determine empirical formula.
↓
*We wil first determine the mass of each element in 1 mol (194.2g) of Caffeine:
C:49.48g C/100g Caffeine x194.2g/mol=96.09gC/mol Caffeine.
H:5.15g H/100g Caffeine x194.2g/mol=10 g H/mol Caffeine.
O:16.49g O/100g Caffeine x194.2g/mol=32.02g O/mol Caffeine.
Now we will convert to mols:
C: 96.09g C/mol Caffeine x1mol C/12.011g C=8g C/mol Caffeine
H: 10 g H/ mol Caffeine x1mol H/1.008g H=9.92mol H /mol Caffeine
N: 56.07g N/mol Caffeine x1mol N/14g N=4.002mol N/ mol Caffeine
O:32.02g O/ mol Caffeine x 1mol O/16g O=2.001mol O/mol Caffeine
C8H10N4O2



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