c12h24
To determine the molecular formula from the empirical formula CFBrO, we first calculate the molar mass of the empirical formula: C (12.01 g/mol) + F (19.00 g/mol) + Br (79.90 g/mol) + O (16.00 g/mol) = 126.91 g/mol. Next, we divide the given molar mass (381.01 g/mol) by the empirical formula mass (126.91 g/mol) to find the ratio: 381.01 g/mol ÷ 126.91 g/mol ≈ 3. The molecular formula is thus (CFBrO)₃, or C₃F₃Br₃O₃.
Molar Mass
The mass of 35 gmoles of carbon monoxide ( CO ) is given by : m = ( n ) ( M ) m = ( 35 gmol ) ( 28.00 g / gmol ) = 980 g <----------------
mass of h = 8 atoms of h/molecule * 1.00794 g/mol = 8.06352g/mol of h in c7h8 mass of c7h8 = 8 atoms of H/molecule * 1.00794 g/mol + 7 atoms/molecule of C * 12.0107 g/mol = 92.13842 g/mol h / c7h8 = 8.06352 / 92.13842 = 0.08752 0.08752 = 8.752% = 9%
C5h10
c12h24
This formula is for magnesium chloride hexahydrate: MgCl2.6H2O.
The molecular formula of the compound is C2H4, which has a molar mass of 28 g/mol. Since the given compound has a molar mass of 42.0 g/mol, it must include an additional CH2 group, resulting in the molecular formula C2H6.
c12h24
The average molecular weight of dry air is 28.96 g/gmol.
The compound's empirical formula is CH3N, and its molecular formula is C4H12N. Therefore, the subscript on C in the chemical formula is 4.
c3h6
The molar mass of the compound is 298 g/mol. To find the molecular formula, we need to calculate the molar mass of Cs and O; Cs has a molar mass of approximately 132.91 g/mol, and O has a molar mass of approximately 16 g/mol. Using these values, we can determine that the compound's molecular formula is Cs11O11.
gmol-1 refers to grams per mole and is a unit of measurement commonly used in chemistry to express molar mass or molecular weight. It represents the mass of one mole of a substance in grams.
The empirical formula of SN has a formula unit mass of the sum of the gram atomic masses of nitrogen and sulfur, i.e., about 46.0667. The gram molecular mass given in the problem divided by this formula unit mass is about 4. Therefore, the molecular formula is S4N4.
To determine the molecular formula from the empirical formula CFBrO, we first calculate the molar mass of the empirical formula: C (12.01 g/mol) + F (19.00 g/mol) + Br (79.90 g/mol) + O (16.00 g/mol) = 126.91 g/mol. Next, we divide the given molar mass (381.01 g/mol) by the empirical formula mass (126.91 g/mol) to find the ratio: 381.01 g/mol ÷ 126.91 g/mol ≈ 3. The molecular formula is thus (CFBrO)₃, or C₃F₃Br₃O₃.