Calculate the mass of 1.000 mole of CaCl(2).
110.986 g/mol
118.56 g/mol
342.15 g/mol
106.323 g/mol
Calculate grams in 3.0000 moles of CO(2).
132.03 g
102.01 g
124.09 g
137.15 g
Calculate number of moles in 32.0 g of CH(4).
1.87 mol
1.99 mol
2.11 mol
2.03 mol
Determine mass in grams of 40.0 moles of Na(2)CO(3).
4,240 g
5,565 g
4,145 g
4,056 g
Calculate the percent composition of KNO(3).
Molar mass = 151.1 g/mol Potassium: (39.10 / 101.1) x 100 = 38.67% Nitrogen: (14.01 / 101.1) x 100 = 13.86% Oxygen: (48.00 / 101.1) x 100 = 12.58%
Molar mass = 164.1 g/mol Potassium: (39.10 / 101.1) x 100 = 38.67% Nitrogen: (14.01 / 101.1) x 100 = 18.96% Oxygen: (48.00 / 101.1) x 100 = 47.48%
Molar mass = 101.1 g/mol Potassium: (39.10 / 101.1) x 100 = 38.67% Nitrogen: (14.01 / 101.1) x 100 = 13.86% Oxygen: (48.00 / 101.1) x 100 = 47.48%
Calculate percent composition for H(2)SO(4).
Molar mass = 98.07 g/mol Hydrogen: (2.016 / 98.07) x 100 = 2.06% Sulfur: (32.06 / 98.07) x 100 = 32.69% Oxygen: (64.00 / 98.07) x 100 = 65.26%
Molar mass = 98.07 g/mol Hydrogen: (2.016 / 98.07) x 100 = 3.09% Sulfur: (32.06 / 98.07) x 100 = 54.12% Oxygen: (64.00 / 98.07) x 100 = 65.39%
Molar mass = 98.07 g/mol Hydrogen: (2.016 / 98.07) x 100 = 9.05% Sulfur: (32.06 / 98.07) x 100 = 44.75% Oxygen: (64.00 / 98.07) x 100 = 14.96%
An oxide of chromium is found to have the following percent composition: 68.4% Cr and and 31.6% O. What is the compound's empirical formula?
CrO(4)
Cr(6)O(2)
Cr(2)O(3)
Cr(5)O
A sample with a molar mass of 34.00 g/mol is found to consist of 0.44g H and 6.92g O. Find its molecular formula.
HO
H(2)O
The percent composition of a compound was found to be 63.5% silver, 8.2% nitrogen, and 28.3% oxygen. Determine the compound's empirical formula.
Ag(3)NO
Ag(2)NO(3)
Ag(4)NO
AgNO(3)
If 4.04g of N combine with 11.46g O to produce a compound with a molar mass of 108.0 g/mol, what is the molecular formula of this compound?
N(3)O(6)
N(2)O(5)
N(1)O(8)
N(2)O(3)
Percent composition Molar mass Molecular formula Empirical formula( Percent composition, Molar mass, Molecular formula, Empirical formula ) is the percent by mass of each element in a compound.
Percent Composition Avogadro's number Molecular Formula Molar Mass( Percent Composition, Avogadro's number, Molecular Formula, Molar Mass ) is the number 6.02 * 10^23, which is the number of representative particles in a mole.
Molar mass Molecular Formula Avogadro's Number Percent Composition( Molar mass, Molecular Formula, Avogadro's Number, Percent Composition ) is the mass in grams of one mole of any pure substance.
molar mass mole hydrate empirical formula( molar mass, mole, hydrate, empirical formula ): the SI base unit used to measure the amount of a substance. It is the number of representative particles, carbon atoms, in exactly 12 g of pure carbon-12.
A hydrate Molar mass A mole Empirical formula( A hydrate, Molar mass, A mole, Empirical formula ) is a compound that has a specific number of water molecules bound to its atoms.
Empirical formula Molecular formula( Empirical formula, Molecular formula ) is a formula that specifies the actual number of atoms of each element in one molecule or formula unit of the substance.
Empirical formula Molecular formula( Empirical formula, Molecular formula ) is a formula that shows the smallest whole number mole ratio of the elements of a compound and may or may not be the same as the empirical formula molecular formula( empirical formula, molecular formula ).
Calculate percent composition of C(2)H(5)OH.
Molar mass = 46.07 g/mol Carbon: (24.022 / 46.07) x 100 = 12.64% Hydrogen: (6.048 / 46.07) x 100 = 13.13% Oxygen: (16.00 / 46.07) x 100 = 94.72%
Molar mass = 55.07 g/mol Carbon: (24.022 / 46.07) x 100 = 22.55% Hydrogen: (6.048 / 46.07) x 100 = 14.13% Oxygen: (16.00 / 46.07) x 100 = 34.73%
Molar mass = 86.07 g/mol Carbon: (24.022 / 46.07) x 100 = 52.14% Hydrogen: (6.048 / 46.07) x 100 = 13.13% Oxygen: (16.00 / 46.07) x 100 = 34.73%