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What is the empirical formula of 0.039 mole of iron atoms combined with 0.052 mole of oxygen atoms?
Wiki User
∙ 12y ago
Divide the larger mole number by the smaller, which is 1
0.052 mole O/0.039 mole Fe
= 1.3333
Using judgment here one calls this number 1
so
FeO
empirical formula and ferrous oxide
Wiki User
∙ 12y ago
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What is the empirical formula of chromium oxide?
Two Chromium Atoms, and 3 Oxygen Atoms. (Cr2O3)
Does no and no2 have the same empirical formulas?
No: The formula NO shows equal numbers of nitrogen and oxygen atoms in the compound, but the formula NO2 shows twice as many oxygen atoms as nitrogen atoms.
How many atoms are present in the formula of ammoium nitrate?
Nine atoms represent the empirical formula of ammonium nitrate. Namely, 2 nitrogen atoms, 3 oxygen atoms and 4 hydrogen atoms.
What is the empirical formula for dinitrogen tetroxide?
Dinitrogen Teroxide = N2O4 Structure: O2N-NO2
How does the empirical formula differ form the molecular formula?
the empirical formula does not include the # of atoms present in an element and the molecular formula does.
What is the empirical formula of chromium oxide?
Two Chromium Atoms, and 3 Oxygen Atoms. (Cr2O3)
A formula that shows only the types and numbers of atoms combined in a single molecule is called?
A formula unit
Does no and no2 have the same empirical formulas?
No: The formula NO shows equal numbers of nitrogen and oxygen atoms in the compound, but the formula NO2 shows twice as many oxygen atoms as nitrogen atoms.
What is the empirical formula weight of HO?
The Empirical formula simply means that it is the lowest ratio of atoms, not the number of atoms. Therefore, because the ratio of Hydrogen atoms to Oxygen atoms is 1:1, the weight of this is the same as normal:1.008 + 16 = 17.008 amu
How many atoms are present in the formula of ammoium nitrate?
Nine atoms represent the empirical formula of ammonium nitrate. Namely, 2 nitrogen atoms, 3 oxygen atoms and 4 hydrogen atoms.
What is the empirical formula for dinitrogen tetroxide?
Dinitrogen Teroxide = N2O4 Structure: O2N-NO2
What is the empirical formula of a compound that contains 57 percent sulfur and 43 percent oxygen by mass?
Since the atomic mass of sulfur is very close to twice the atomic mass of oxygen, the ratio of numbers of oxygen atoms to numbers of oxygen atoms is close to 43:28.5. The closest ratio of small integers is 3:2; therefore the empirical formula is S2O3.
How does the empirical formula differ form the molecular formula?
the empirical formula does not include the # of atoms present in an element and the molecular formula does.
Empirical and molecular formulas?
The Empirical formula of Al2Br6 is AlBr3.
What is the empirical formula of a compound that contains 0.20 moles of nitrogen atoms and 0.40 moles of oxygen atoms?
Divide the larger by the smaller mole number. The smaller mole number is assumed to be 1. 0.40/0.20 = 2 As nitrogen is 0.20 moles we have 1 atom of it and 2 atoms of oxygen, so NO2 is the empirical formula (nitrogen dioxide).
What is the empirical formula for the compound below C10H25O5?
C2H5O. Empirical formula is the simplest integer ratio of atoms.
How do you do an empirical formula?
An empirical formula is a chemical formula which contain only the chemical symbol of elements; no number of atoms or structure.
