Question

Please I really really need help!!!

1) If there are 0.375 moles 15 grams of compound, what would its molar mass be?
2) How many moles are there in 90 grams of magnesium oxide?
3) Calculate the molecular mass of ammonia.
4) Calculate the formula mass of sodium hydrogen sulphate.
5) Assuming the relation atomic mass of metal M to be 56, what would the empirical formula of its oxide containing 70.0% of M be?
I want you to answer all of the questions, 5 of them!!!!
And don't answer the question just for points if you don't know what it means!!!!
And also show workings I won't accept an answer without an explanation!!!!​

Answer 1

1) 40

2) 2.25 moles

3) 17

4) 120

5) Fe₂O₃

Step-by-step explanation:

Please see attached picture for full solution.

Answer 2

1)
`\boxed{Molar \ mass = 40\ g/mol}`

2)
`\boxed{No.\ of\ Moles = 1.61 \ moles}`

3)
`\boxed{Ammonia = 16\ g/mol}`

4)
`\boxed{Sodium\ Hydrogen\ Sulphate= 120 \ g/mol}`

5)
`\boxed{Fe_(2)O_(3)}`

Step-by-step explanation:

Question # 1:

Using Formula, No. of moles = Mass in grams / molar mass

Where moles = 0.375 mol. , mass = 15 g

=> Molar Mass = Mass in grams / No. of moles

=> Molar Mass = 15 / 0.375

=> Molar mass = 40 g/mol

Question # 2:

No. of moles = Mass in grams / Molar mass

Where Mass = 90 g ,

=> Molar Mass =
`MgO_(2)`

=> MM = 24 + (16*2)

=> MM = 24 + 32

=> MM = 56 g/mol

No. of Moles = 90 / 56

No. of Moles = 1.61 moles

Question # 3:

Ammonia =>
`NH_(3)`

N has atomic mass 14 ang H has atomic mass 1

Ammonia = (14)+(1*2)

Ammonia = 14+2

Ammonia = 16 g/mol

Question # 4:

Sodium Hydrogen Sulphate =>
`NaHSO_(4)`

Where Na has atomic mass 23, H has atomic mass 1 , sulpher 32 and oxygen 16

`NaHSO_(4)` = 23+1+32+(16*4)

=> 56+64

=> 120 g/mol

Question # 5:

The metal which has relative atomic mass of 56 is Iron (Fe)

Given that the oxide contains 70.0 % of Metal

Mass of
`MO_(x)` (Metal oxide) = 56 / 70 * 100

Mass of
`MO_(x)` (Metal oxide) = 0.8 * 100

Mass of
`MO_(x)` (Metal oxide) = 80

Mass of x O's = 80 - 56

Mass of x O's = 24

Now, Mass of O = 24 / 16

Mass of O = 1.5

So, Fe Metal and its oxide are in the ratio of

1 : 1.5

×2 ×2

2 : 3

So, the empirical formula of metal with its oxide is
`Fe_(2)O_(3)`