Empirical formula is the simplest ratio of whole numbers of components in a compound
calculating for 100 g of compound
C H O
mass 64.27 g 7.19 g 28.54 g
number of moles 64.27 g / 12 g/mol 7.19 g/1 g/mol 28.54 g / 16 g/mol
= 5.356 mol = 7.19 mol = 1.784 mol
divide by least number of moles
5.356 / 1.784 7.19 / 1.784 1.784 / 1.784
= 3.002 4.03 = 1.000
rounded off to nearest whole number
C - 3
H - 4
O - 1
empirical formula - C₃H₄O
mass of empirical formula = 12 g/mol x 3 + 1 g/mol x 4 + 16 g/mol x 1 = 56 g
molecular mass = 168.19 g/mol
molecular formula is the actual ratio of elements making up the compound
number of empirical units = molar mass of molecule / empirical mass
empirical units = 168.19 g/mol / 56 g = 3.00
there are 3 empirical units making up the molecular formula
molecular formula = 3 x C₃H₄O
molecular formula = C₉H₁₂O₃
Answer:
The products will be;
CO₂ + H₂O + NaC₂H₃O₂
Explanation:
We are given;
Two reactants NaHCO₃ and HC₂H₃O₂
We are supposed to determine the products;
- We need to know that hydrogen carbonates reacts with acids to give water, carbon dioxide and a salt as the products.
Therefore;
- In this case, sodium hydrogen carbonates (baking soda) reacts with acetic acid to form water, carbon dioxide and sodium acetate.
- The equation for the reaction is;
NaHCO₃ + HC₂H₃O₂ → CO₂ + H₂O + NaC₂H₃O₂
- Therefore, the products of the complete reaction between sodium hydrogen carbonate and acetic acid are CO₂ + H₂O + NaC₂H₃O₂
Answer:
0.454moles
Explanation:
Calculate the molar mass of AgBr
Ag=108
Br=80
Molar mass= 108+80
Molar mass=188g/mol
It's mole to gram question
2moles of Na2S2O3= 188g of AgBr
X moles of Na2S2O3=42.7g of AgBr
Cross multiply
188x=2×42.7
X= 85.4/188
X=0.454moles
Answer:
1.- Chemical change
2.- Because the atoms in the image are tranformed into a new molecule
3.- Yes, it does
4.- Due to the amount of atoms are the same both in products and reagents
Explanation:
