Answer:
Molecular formula = C₆H₁₂O₆
Explanation:
Given data:
Mass of hydrogen = 31.7 g
Mass of carbon = 283.4 g
Mass of oxygen = 377.4 g
Molar mass of compound = 176.124 g/mol
Molecular formula = ?
Solution:
Number of gram atoms of H = 31.7 / 1.01 = 31.4
Number of gram atoms of O = 377.4 / 16 = 23.6
Number of gram atoms of C = 283.4 / 12 = 23.6
Atomic ratio:
C : H : O
23.6/23.6 : 31.4/23.6 : 23.6/23.6
1 : 1.33 : 1
C : H : O =3 (1 : 1.33 : 1
)
Empirical formula is C₃H₄O₃.
Molecular formula:
Molecular formula = n (empirical formula)
n = molar mass of compound / empirical formula mass
Empirical formula mass = 3×12+4+3×16 = 88
n = 176.124 / 88
n = 12
Molecular formula = n (empirical formula)
Molecular formula = 2 (C₃H₄O₃)
Molecular formula = C₆H₁₂O₆
Answer:
The answer is: 51.8 g (86% of serving size)
Explanation:
In order to solve the problem, we have to first determine the number of moles there are in 11.0 g of sucrose. Sucrose has a molecular weight of 342 g (we calculate this from the molar mass of the elements : 12 x 12 g/mol C + 22 x 1 g/mol H + 11 x 16 g/mol O). So, we divide the mass (11.0 g) into the molecular weight of sucrose:
11.0 g sucrose x 1 mol/342 g sucrose= 0.032 mol
We have 0.032 mol of sucrose in a serving of 60 g. But we need less moles (0.0278 mol):
0.032 mol ------------ 60 g serving
0.0278 mol------------ x= 0.0278 mol x 60 g serving/0.032 mol
x= 51.8 g
So, lesser than 1 serving of 60 g must be eaten to consume 0.0278 mol os sucrose. Exactly, 51.8 g (which stands for a 86% of the serving size).
D has a total of four significant figures.
Answer:
39.2 %
Explanation:
The following data were obtained from the question:
Mass of sample = 24 g
Mass of Cl = 14.6 g
Mass of B = 9.4 g
Percentage composition of boron =?
The percentage composition (by mass) of boron in the sample can be obtained as illustrated below:
Percentage composition of boron = mass of B /mass of sample × 100
Percentage composition of boron = 9.4/24 × 100
Percentage composition of boron = 39.2 %
Therefore, the percentage composition (by mass) of boron in the sample is 39.2 %
Answer:
Molar mass = 3.9236 g/mol ≅ 4 g/mol
This corresponds to Helium gas.
Explanation:
Let the moles of nitrogen gas = x moles
Moles of carbon dioxide = x moles ( As both are filled at same temperature and pressure conditions )
Given:
Molar mass of nitrogen gas, 
= 28.014 g/mol
The formula for the calculation of moles is shown below:
Thus,
Mass of nitrogen gas = 28.014x g
So,
Let, 
Similarly,
Molar mass of nitrogen gas, 
= 44.01 g/mol
The formula for the calculation of moles is shown below:
Thus,
Mass of nitrogen gas = 44.01x g
So,
Solving the two equations, we get :
x = 0.00943 moles
Thus, Given:
Mass of the gas = 0.037 moles
Moles = 0.00943 moles
The formula for the calculation of moles is shown below:
Thus,
Molar mass = 3.9236 g/mol ≅ 4 g/mol
This corresponds to Helium gas.
