The chemical compound's empirical formula is NS.
The chemical compound's molecular formula is N4S4.
<h3>What does a chemical empirical formula look like?</h3>
- The empirical formula of a compound that gives the proportion (ratios) of the elements in the complex but not the precise number or arrangement of atoms is known as an empirical formula.
- This would be the compound's element to whole number ratio with the lowest value.
<h3>What sort of empirical formula would that be?</h3>
- The chemical structure of glucose is C6H12O6. Every mole of carbon and oxygen is accompanied by two moles of hydrogen.
- Glucose has the empirical formula CH2O.
- Ribose has the chemical formula C5H10O5, which can be simplified to the empirical formula CH2O.
learn more about empirical formula here
brainly.com/question/1603500
#SPJ4
the question you are looking for is
A compound containing only sulfur and nitrogen is 69.6% S by mass; the molar mass is 184 g/mol. What are the empirical and molecular formulas of the compound?
Answer:
No element shares the same atomic number. The atomic number of every element is unique to itself, just like a fingerprint is unique to a human. The atomic number of element allows us to identify it, like how we can identify a person from their fingerprint.
Explanation:
Answer:
Sorry for disrespecting you but i don't know your language. I want to help but my problem is your language IM SO SORRY
Answer:
Molecular formula
Explanation:
Molecular formula in the first place is required to understand which compound we have. We then should refer to the periodic table and find the molecular weight for each atom. Adding individual molecular weights together would yield the molar mass of a compound.
Then, dividing the total molar mass of a specific atom by the molar mass of a compound and converting into percentage will provide us with the percentage of that specific atom.
E. g., calculate the percent composition of water:
- molecular formula is ;
- calculate its molar mass: [tex]M = 2M_H + M_O = 2\cdot 1.00784 g/mol + 16.00 g/mol = 18.016 g/mol;
- find the percentage of hydrogen: [tex]\omega_H = \frac{2\cdot 1.00784 g/mol}{18.016 g/mol}\cdot 100 \% = 11.19 %;
- find the percentage of oxygen: [tex]\omega_O = \frac{16.00 g/mol}{18.016 g/mol}\cdot 100 \% = 88.81 %.