1.Calculate the mass of carbon in grams, the percent carbon by mass, and the num
ID: 972014 • Letter: 1
Question
1.Calculate the mass of carbon in grams, the percent carbon by mass, and the number of individual carbon atoms present in 1.66×1021 molecules of carbon monoxide.?
2.Calculate the mass of carbon in grams, the percent carbon by mass, and the number of individual carbon atoms present in 6.365 g of carbon suboxide, C3O2.?
3.Calculate the mass in grams of 1.5×1010 nitrogen molecules?
4.Calculate the number of atoms of each element present in 0.000392 g of benzene, C6H6 ?
5.Calculate the number of atoms of each element present in 7.13 g of carbon dioxide?
6.Calculate the number of atoms of each element present in 3.92 g of water?
7.An organic acid is composed of carbon (58.80%), hydrogen (9.89%), and oxygen (31.33%). Its molar mass is 102.13 g/mol. Determine the molecular formula of the compound.
8.Quinine, C20H24N2O2 is the bitter tasting compound in tonic water. It is also the compound used to prevent malaria. How many atoms of hydrogen are in 1.0 mol of quinine?
9.Dimethylnitrosamine (CH3)2N2O is a carcinogenic (cancer causing) substance that may be formed in foods, beverages or gastric juices from the reaction of nitrate ion (used as a food preservative) with other substances.
10.What is the mass of 1.0 × 1015 molecules of dimethylnitrosamine?
11.Vitamin B12, cyanocobalamin, is essential for human nutrition.
Its molecular formula is C63H88CoN14O14P. A lack of this vitamin in the diet can lead to anemia. Cyanocobalamin is the form of the vitamin found in vitamin supplements.
What is the mass (in grams) of one molecule of cyanocobalamin?
12.How many atoms of carbon are present in 1.0 g of CH2Cl2?
13.How many atoms of carbon are present in 1.0 g of C12H22O11?
14.How many atoms of nitrogen are present in 25.0 g of CO(NH2)2?
15.The chemical formula for vitamin E is C29H50O2. What is the mass percent for each element in one mole of vitamin E?
What is the mass percent for each element in 1 mole of strontium nitrate?
16.A compound with molar mass 180.1 g/mol has the following composition:
Determine the empirical formula ?
17.A compound with molar mass 180.1 g/mol has the following composition:
Determine the molecular formula
(Express your answer as a chemical formula.)
18. 1.6 mg of chromium (the average atomic mass for chromium atoms is 52.00 amu)?
19.1.6e-3 g of strontium (the average atomic mass for strontium atoms is 87.62 amu)?
20. 1.61e4 g of boron (the average atomic mass for boron atoms is 10.81 amu) ?
21.
1.6e-6 g of californium (the average atomic mass for californium atoms is 251 amu)?
22. 1.2 ton (2400 lb) of iron (the average atomic mass for iron atoms is 55.85 amu)?
C 40.0% H 6.70% O 53.3%Explanation / Answer
You should try to give 4-5 questions each time.
1) 1.66×1021 molecules of carbon monoxide
= 1.66×1021 / 6.023*1023
= 0.00276 moles of CO
One mole of CO contains 12 g C.
mass of carbon in grams = 0.00276*12 = 0.033 g
percent carbon by mass = [0.033/ (0.00276 * molar mass of CO)]*100
= 0.033*100/0.00276*28
= 42.7%
number of individual carbon atoms present
= 0.00276 * 6.023*1023
= 1.66×1021
2. 6.365 g of carbon suboxide, C3O2
= 6.365 g /[3*12 + 2*16]
= 0.0936 moles of C3O2
One mole of C3O2 contains 36 g C .
mass of carbon in grams = 0.0936 * 36 = 3.37 g
percent carbon by mass
= (3.37/6.365)*100
= 52.9%
number of individual carbon atoms present
= 0.0936 * 3 * 6.023*1023
= 1.69×1023
3. mass in grams of 1.5×1010 nitrogen molecules
= (1.5×1010 / 6.023*1023) * 28 g
= 6.97*10-13 g
4. 0.000392 g of benzene,
= 0.000392/78 moles
= 5*10-6 mole
No of C atoms = 5*10-6 * 6 * 6.023*1023
= 1.816 *1019
No of H atoms = 5*10-6 * 6 * 6.023*1023
= 1.816 *1019
5 and 6) same procedure as shown in the previous question.
7. molar mass is 102.13 g/mol
mass of C in one mole of the compound = 102.13 g * 58.8%
= 60 g
No of C atoms in one formula unit = 60/12 = 5
mass of H in one mole of the compound = 102.13 g * 9.89%
= 10 g
No of H atoms in one formula unit = 10/1 = 10
mass of O in one mole of the compound = 102.13 g * 31.33%
= 32 g
No of O atoms in one formula unit = 32/16 = 2
molecular formula of the compound = C5H10O2 or CH3CH2CH2CH2COOH
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