المجموعه السابعه | اسئلة واجب الكيمياء وحلها للي ماقدرت تاخذهم ,
 

Assignment
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?
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.


.We will first determine the mass of each element in 1 mole (194.2 g) of caffeine,
Mass of C = 49.48/100 x 194.2 = 96.09 g/mol C
Mass of H = 5.15/100 x 194.2 = 10.0 g/mol H
Mass of N = 28.87/100 x 194.2 = 56.07 g/mol N
Mass of O = 16.49/100 x 194.2 = 32.02 g/mol O
Now we will convert to moles,
No of moles of C = 96.09/12.01 = 8.001 mol C/mol caffeine.
No of moles of H = 10.0/1.008 = 9.92 mol H/ mol caffeine.
No of moles N = 56.07/14.01 = 4.002 mol N / mol caffeine.
No of moles of O = 32.02/16.00 = 2.001 mol O/ mol Caffeine.
the molecular formula of caffeine can be given as: C8H10N4O2

رد: المجموعه السابعه | اسئلة واجب الكيمياء وحلها للي ماقدرت تاخذهم ,
 

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