Respuesta :
Answer:
1) empirical formula = CH2O ; the molecular formula = C2H402
2) empirical formule = C3H404 = Molecular formula
Explanation:
CxHyOz → CO2 + H2O
⇒ So for CO2 we can say that for each 1 C-atom (12g) there are 2 O-atoms (32g).
This means the ratio is 12g C/ 32g O
CO2 has a molar mass of 44 g/mole
in 14.57 grams of CO2 there is: (12g Carbon / 44g CO2) * 14.57 g CO2 = 3.9736 g Carbon
⇒For H2O we can do the same:
The ratio of each element in H2O is: 2 H-atoms/ 1 O-atom This is 2g H/ 16g O
In 5.966 g of water there are: [2 g H / 18 g H2O] * 5.966 g H2O = 0.6629 g H
⇒The original sample had 9.941 g of Sample - 3.9736 g of C - 0.6629 g of H = 5.3045 g of O
2) Calculate number of moles
C: 3.9736 g / 12.0 g/mol = 0.3311 mol
H: 0.6629 g / 1.0 g/mol = 0.6629 mol
O: 5.3045g / 16.0 g/mol = 0.3315 mol
3) Now we should divide each number of mole by the smallest number (0.3311) to find the proportion of the elements
C: 0.3311/ 0.3311= 1
H: 0.6629 / 0.3311 = 2
O: 0.3315/ 0.3311 = 1
This gives us the empirical formula of CH2O
4) Calculate the mass of the empirical formula
Molar mass of Carbon = 12g /mole
Molar mass of Hydrogen = 1g /mole
Molar mass of Oxygen = 16g /mole
mass of the empirical formule = 12 + 2*1 + 16 = 30g
5) Calculate the molecular formula
mass of molecular formule / mass of empirical formula = n
We have to multiply the empirical formula by n to get the molecular formula.
60 / 30 = 2 = n
If we multiply CH2O by 2 we'll get: C2H4O2
If we control this by calculating the molar mass:
2*12 + 4*1.01 + 2*16 = 60.04 g/mole
Then, the molecular formula is C2H4O2
CxHyOz → CO2 + H2O
⇒ So for CO2 we can say that for each 1 C-atom (12g) there are 2 O-atoms (32g).
This means the ratio is 12g C/ 32g O
CO2 has a molar mass of 44 g/mole
in 17.97 grams of CO2 there is: (12g Carbon / 44g CO2) * 17.97 g CO2 = 4.90 g Carbon
⇒For H2O we can do the same:
The ratio of each element in H2O is: 2 H-atoms/ 1 O-atom This is 2g H/ 16g O
In 4.905 g of water there are: [2 g H / 18 g H2O] * 4.905 g H2O = 0.545 g H
⇒The original sample had 14.16 g of Sample - 4.90 g of C - 0.545 g of H = 8.715g of O
2) Calculate number of moles
C: 4.90 g / 12.0 g/mol = 0.4083 mol
H: 0.545 g / 1.0 g/mol = 0.545 mol
O: 8.715g / 16.0 g/mol = 0.5447 mol
3) Now we should divide each number of mole by the smallest number (0.4083) to find the proportion of the elements
C: 0.4083/ 0.4083= 1
H: 0.545 / 0.4083 =1.33
O: 0.5447/ 0.4083 = 1.33
We should multiply everything by 3
This gives us the empirical formula of C3H4O4
4) Calculate the mass of the empirical formula
Molar mass of Carbon = 12g /mole
Molar mass of Hydrogen = 1g /mole
Molar mass of Oxygen = 16g /mole
mass of the empirical formule = 3*12 + 4*1 + 4*16 = 104
5) Calculate the molecular formula
mass of molecular formule / mass of empirical formula = n
We have to multiply the empirical formula by n to get the molecular formula.
104 / 104 = 1 = n
This means the empirical formula = molecular formula = C3H4O4
