How to calculate "Empirical Formula" || Super Trick Method with Q&A | Mole concept || NEET JEE

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If you really want to know the trick to how to find emprical formula when percentage is given then start from 12:35

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thanks really time is very precious

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Why didn't I see your comment . 😞😞

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[@anandshete9170]

%composition of H in H2O is 11.11 not 11.79
and %composition of O is 88.88 not 88.21 according to your calculations

[@alphaofficial914]

Yes man I am also confused

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Sir ji💪💓

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Mera bhi 🙋🙋

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Itna hi pta ah too yaha kyu aaya ha

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You are right Bro
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Hello Good evening sir. 😄😄🌹🌹🌷🌷🌺🌺🌺
Lecture completed 👍👍

[@anjleenkaur65]

Good evening sir. 😊 🙏
Thanks for the lecture.. you are the best

[@owaisosman6703]

Thanks allot sir and made ejee team going to complete this....

[@shyamsarkar7994]

Sir
In C2H5OH
% of carbon =52.17
% of hydrogen =13.04
% of oxygen = 34.78

[@ankitachaturvedi5994]

Kse smjh ni aaya 24 carbon or 6 hydrogen kse kha se aaya??

[@Jethalalsigma231]

@@ankitachaturvedi5994 Isme carbon ke 2 atom hai 1 atom ka mass 12 two atom = 12×2 = 24
Hydrogen ke 6 atom
1 atom ka mass = 1 U 1U × 6 = 6u

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[@dhruvdhruv9496]

sir please tell me how to solve the question " determine the emperical formula of an oxide of iron which has 69.9% iron and 30.1% dioxygen by mass

[@sanjuverma2539]

Its an Example in Book

[@FunTime-tg7ki]

Take notes

[@unknownxyz2353]

Bhai Tera ans aayega fe2o3

[@unknownxyz2353]

Phele 69.9 ko approx 70 g = given mass man then divide by the atomic mass of fe that's 56 u will get 70 /56 = 20/16 or 5/4 . Now take 30.1 as given mass for oxygen / divide this by 16 g which is the atomic mass for oxygen . So u will get 30 /16 here I have approxed 30.1 to 30 now since 20 /16 is smaller divide 30 /16 by it and u will get 3 :2 = moles of o :moles of fe . Compare and get the formula Fe2O3 and also mf = ef ( Here the simplest ratio is 2:3 itself )

[@PUBG-PLAYER.-]

Fantastic video i got everything related to E.F and M.F. thnku sir🙏🙏🙏👍👍☺

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Allah aapko salamat rakhe hamesha..with ur family...u r helping the students..Allah will never let u down..anywhere...insha allah

[@humzashaikh4041]

Thank u Sir ur the best teacher of chemistry. ....😊
Answer c=52.1%
H=13.4%
O=34.7%👍👍👍

[@poojarajput4481]

H 13.04 hoga

[@anjalimishra4192]

13.04%

[@manavthevar2132]

Yes H is 13.04 but still good you tried and succeeded

[@prakharupadhyay4755]

Sir you are the best teacher of chemistry I see ever your videos clear all doubts very easily thank you sir 🙏🙏🙏🙏

[@sirvilive2935]

Completed......
Thankyou so much Sir ji

[@akankshalakhera3774]

C=52.17% ,H=13.04,O=34.34%
Thank you so much sir 👍👍👍👍👍👍👍👍👍👍👍👍

[@Utkarsh_Singh_Hrithik2024]

Sir from where U R
Such a Great 👍🏻 Explanation 😃
First time I understand It
Everything 😊

[@Ruhi1919]

C℅=52.17
%H=14.60
%O=34.7
These are % composition of C2H5OH

[@physicschemistry7448]

Percentage of hydrogen is wrong

[@physicschemistry7448]

Its 13.04

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In a day two video. Just sufficient. Still want more😍😍

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Understanding with clarity.we love u and ur advice.hope I fulfill my dream to crack jee watching u.searching for good books from which I can prepare jee in few months

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bhai hoagyi kya tumhari NEET/JEE ??

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I am gonna share this vdo to all my friends they will also get cleared this concept🔥😄

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Sir mujhe pata nhi tha ki aap itne acche teacher bhi ho

[@meeta_ramaiya]

Good evening sir Thank-you for lecture👍👍👍C=52.17% H=13.04% O=34.78%

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Thanks sir finally i choose a better platform ☺️☺️ thanks a lots sir

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Absolutely fine lecture for easy understanding thanks for such a excellent video.❤️❤️

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thank you sir apki help se mera chemistry bhut strong ho gya h..

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You are really a good teacher of chemistry

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Good evening sir😄😄

[@Its_beauty_of_nature]

10;53
percent of carbon=24/46x100=52.21percent
percent of hydrogen=6/46x100=12.12percent
percent of oxygen=16/46x100=35.67percent....

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Ek no explanation❤❤
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My mole concept is completely clear because of you.Thanks sir😊😊☺️☺️

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Bhagwaan aap jaisa log is dharti par bnate rahe.

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Sir ye second video h jo Mene dekhi h ap ne boot ache se samjhaya h 👌👍

[@pareihanbasagapam6500]

How about rounding up to their simplest whole number in case of decimal.. Plz explain

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Yes plz make a video on this. Its creating a big problem for us 🙏

[@InDrayanI_Chavhan]

Completed Sir....👍🏻👍🏻👍🏻👍🏻
Ans is...C = 52.17%, H = 13.04%,
O = 34.78%
Thanku Sir...🙏🏻🙏🏻🙏🏻😊

[@darshikakushwaha7644]

Indrayani Chavhan Give me ans....moles concept kha h...NCERT me...

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Some basic concept of chemisry
In 11th NCERT..👍🏻👍🏻

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good

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[@ruthwik1933]

Sir if molar mass is not given means how should we calculate
Nice explanation sir
Thank you sir

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No

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Osm video Maine bhut jagah dekha par kahi sahi explained nahi kiya but you are best👍👍

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Hi

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Arvind sir is my 2nd favourite teacher after mkt sir iit explains

[@1MinTakeAway]

Mass percentage of C - 26.08 percentage
H- 13.04 percentage
O- 34.78 percentage

[@1MinTakeAway]

Mass percentage of hydrogen in water is 11.11percentage not 11.79 percentage

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[@riyazmalik6572]

% of carbon = 52.17%
% of hydrogen = 13.04%
% of oxygen = 34.78%

[@rajeshvaid3141]

c-52.17,H-13.04 and o-34.78

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[@meer-e-karvaan.6070]

ans:) C=52.17
H=13.04
O=34.78

[@SimranSingh-hl6sj]

Sir woh C2H8O2 wala direct percentage se bhi nikal sakte hai na toh itna lamba way se kyu kiya aapne??
For example. For carbon. 37.5 * 64 / 100 = 24 and carbon ka mass is 12 toh 2 atoms hue...similarly sabka aise nikal sakte hai na???

[@NawabAsfaarbloch]

Bhut Acha smjhia Hy sir ap na bhut easy way ma❤

[@dadaraonathbhajan]

Ans is c=52.17 h=13.04 o=34.78 and thanks sir

[@persever365]

Percent composition of Carbon= 52.1%
Percent composition of Oxygen= 13.3%
Percent composition of Hydrogen= 13.3%

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Nice explained sir🤩

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best . each word explained nicely

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[@Nunnu-lp7wt]

Sir we can get the mf by converting% into atomic mass
Eg:- C -37.5% = 24g = 2C
H -12.5% = 8g = 8H
O - 50% = 32g = 2O

[@manojlambat6578]

Thanks sir you maked me Free of confusion......................!😊

[@SWETAKUMARI-xr8ox]

C-52;17
H-13;04
O-34 present

[@kanishkmeena6737]

51.17
13.04
34.78
Thank you so much sir 😊😊😊😊
Love you sir 😘😘 so mucha

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Thanks sir aaj pata chala h empirical formula ka important or iska pata chalte h class karne mai or maza aya h 🙃🙃🙃 thanks sir 😘😘😘

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Sir u r the best....sir plz aldehyde ketones chap upload kijiye sir plzz

[@siddhideshmane7950]

% of carbon= 52.17 %
% of hydrogen= 13.04%
% of oxygen= 34.78
in acetic acid

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Thanks for the solution 🙏👍👍

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